Coordinate Geometry Part2

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Exercise 4

Find the equation of the locus of $P\,$ , if $P\,$ moves is such a way that:

a) $AP=2PB\,$ , where $A\left(2,0\right)$ and $B\left(0,3\right)$ .

$3x^{2}+3y^{2}+4x-24y+32=0\,$
- $\begin{align} & \mbox{Let }P\left(x,y\right)\\ & AP=2PB \\ & \sqrt{\left(x-2\right)^{2}+y^{2}}=2\sqrt{x^{2}+\left(y-3\right)^{2}}\\ & \left(x-2\right)^{2}+y^{2}=4\left[x^{2}+\left(y-3\right)^{2}\right]\\ & x^{2}-4x+3+y^{2} = 4\left(x^{2}+y^{2}-6y+9\right)\\ & x^{2}-4x+3+y^{2} =4x^{2}+4y^{2}-24y+36 \\ & 3x^{2}+3y^{2}+4x-24y+32=0 \\ \end{align}$

b) its distance from the point $\left(2,-3\right)$ is $5\,$ units.

$x^{2}+y^{2}-4x+6y-12=0\,$
- $\begin{align} & \mbox{Let }A\left(2,-3\right), P\left(x,y\right)\\ & AP=5 \\ & \sqrt{\left(x-2\right)^{2}+\left(y+3\right)^{2}}=5\\ & x^{2}-4x+4+y^{2}+6y+9 = 25\\ & x^{2}+y^{2}-4x+6y-12=0 \\ \end{align}$

c) it is equidistant from the points $\left(-1,4\right)$ and $\left(2,7\right)$ .

$x+y-6=0\,$
- $\begin{align} & \mbox{Let }A\left(-1,4\right),B\left(2,7\right), P\left(x,y\right) \\ & AP=BP \\ & \sqrt{\left(x+1\right)^{2}+\left(y-4\right)^{2}}=\sqrt{\left(x-2\right)^{2}+\left(y-7\right)^{2}}\\ & x^{2}+2x+1+y^{2}-8y+16 =x^{2}-4x+4+y^{2}-14y+49 \\ & 6x+6y-36=0 \\ & x+y-6=0 \end{align}$

d) it is equidistant from the points $\left(1,3\right)$ and the $x\,$ -axis.

$x^{2}-2x-6y+10=0\,$
- Media:MT-CG-E4-4d.dia
- $\begin{align} & \mbox{Let } A\left(1,3\right), P\left(x,y\right) \\ & AP =d \\ & \sqrt{\left(x-1\right)^{2}+\left(y-3\right)^{2}}=\left|y\right| \\ & \left(x-1\right)^{2}+\left(y-3\right)^{2} = y^{2} \\ & x^{2}-2x+1+y^{2}-6y+9=y^{2} \\ & x^{2}-2x-6y+10=0 \\ \end{align}$

e) $\angle APB=90^{\circ}$ where $A\left(3,-2\right)$ and $B\left(1,4\right)$ .

$x^{2}+y^{2}-4x-2y-5=0\,$
- $\begin{align} & \mbox{Let } P\left(x,y\right) \\ & \angle APB=90^{\circ}, \therefore AP \perp BP \\ & \therefore m_{AP}m_{BP}=-1 \\ &\left(\frac{y+2}{x-3}\right)\left(\frac{y-4}{x-1}\right) = -1 \\ & y^{2}-2y-4 = -\left(x^{2}-4x+3\right)\\ & x^{2}+y^{2}-4x-2y-5=0\\ \end{align}$

f) it is equidistant from the lines $3x-4y+2=0\,$ and $5x+12y-7=0\,$

$14x-112y+61=0 \mbox { or } 64x+8y-9=0 \,$
- $\begin{align} & \mbox{Let } P\left(x,y\right) \\ & d_{1}=d_{2} \\ & \frac{\left|3x-4y+2\right|}{\sqrt{3^{2}+4^{2}}}=\frac{\left|5x+12y-7\right|}{\sqrt{5^{2}+2^{2}}}\\ & 13\left|3x-4y+2\right|=5\left|5x+12y-7\right|\\ & \therefore 39x-52y+26=25x+60y-35 \mbox{ or } 39x-52y+26=-\left(25x+60y-35\right)\\ & 14x-112y+61=0 \mbox { or } 64x+8y-9=0 \\ \end{align}$

g) it’s distance from the point $\left(2,0\right)$ is two times its distance from the line $x=4\,$

$3x^{2}-y^{2}-36x+60=0\,$
- $\mbox{Let } A\left(2,0\right), P\left(x,y\right)$
- $\begin{align} & \left(x,y\right), x+0y-4=0 :\\ & d= \frac{\left|x-4\right|}{\sqrt{1^{2}+0^{2}}}\\ & \quad =\left|x-4\right| \end{align}$ or Media:MT-CG-E4-4g.dia
- $\begin{align} & AP =d \\ & \sqrt{\left(x+2\right)^{2}+\left(y-3\right)^{2}}=2\left|x-4\right| \\ & x^{2}+4x+4+y^{2} = 4\left(x^{2}-8x+16\right) \\ & x^{2}+4x+4+y^{2} = 4x^{2}-32x+64) \\ & 3x^{2}-y^{2}-36x+60=0 \\ \end{align}$

h) it is equidistant from origin and the line $x+y+1=0\,$ .

$x^{2}+y^{2}-2xy-2x--2y-1=0 \,$
- $\begin{align} & \mbox{Let } P\left(x,y\right) \\ & OP =d \\ & \sqrt{x^{2}+y^{2}}=\frac{\left|x+y+1\right|}{\sqrt{1^{2}+1^{2}}} \\ & 2\left(x^{2}+y^{2}\right) = \left(x+y+1\right)\left(x+y+1\right) \\ & 2x^{2}+2y^{2} = x^{2}+2xy+2x+2y+1+y^{2} \\ & x^{2}+y^{2}-2xy-2x--2y-1=0 \end{align}$

i) it is equidistant from the lines $y=2x-1\,$ and $y=2x+6\,$ .

$4x-2y+5=0 \,$
- $\begin{align} & \mbox{Let } P\left(x,y\right) \\ & y=2x-1 \qquad\qquad\quad y=2x+6 \\ & \therefore 2x-y-1=0 \qquad 2x-y-6=0 \\ & d_{1} =d_{2} \\ & \frac{\left|2x-y-1\right|}{\sqrt{2^{2}+1^{2}}}=\frac{\left|2x-y+6\right|}{\sqrt{2^{2}+1^{2}}} \\ & \left|2x-y-1\right| = \left|2x-y-6\right| \\ & 2x-y-1 = 2x-y-6 \mbox{ or } 2x-y-1 = -\left(2x-y-6\right)\\ & -1=6 \mbox{ (rejected)} \qquad\qquad 4x-2y+5=0 \\ & \therefore 4x-2y+5=0 \end{align}$

Exercise 5

1) Find the points of intersections between the following curves/lines. State clearly those cases where they touch.

a) $y=3x^{2},y=2-x\,$

$\left(\frac{2}{3},\frac{4}{3}\right) \mbox{ and }\left(-1,3\right)$
- $\begin{align} & y=3x^{2}\frac{\qquad}{}(1)\\ & y=2-x\frac{\qquad}{}(2)\\ & (1)\to(2):\\ & 3x^{2}=2-x \\ & 3x^{2}+x-2=0 \\ &\left(3x-2\right)\left(x+1\right)=0 \\ & x=\frac{2}{3} , x=-1 \\ & \mbox{When } x=\frac{2}{3}, y=\frac{4}{3}; x=-1, y=3 \\ & \therefore \left(\frac{2}{3},\frac{4}{3}\right) \mbox{ and }\left(-1,3\right) \end{align}$

b) $y=4x^{2}-x,y=3x-1\,$

$\therefore \left(\frac{1}{2},\frac{1}{2}\right)\mbox{ (touch)}$
- $\begin{align} & y=4x^{2}-x\frac{\qquad}{}(1)\\ & y=3x-1\frac{\qquad}{}(2)\\ & (1)\to(2):\\ & 4x^{2}-x=3x-1 \\ & 4x^{2}-4x+1=0 \\ &\left(2x-1\right)^{2}=0 \\ & x=\frac{1}{2} \mbox{ (repeated)}, y=\frac{1}{2} \\ & \therefore \left(\frac{1}{2},\frac{1}{2}\right)\mbox{ (touch)} \end{align}$

c) $y=1-2x^{2},y=5-x\,$

$\mbox{ no points of intersection}\,$
- $\begin{align} & y=1-2x^{2}\frac{\qquad}{}(1)\\ & y=5-x\frac{\qquad}{}(2)\\ & (1)\to(2):\\ & 1-2x^{2}=5-x \\ & 2x^{2}-x+4=0 \\ & b^{2}-4ac = 1-4\left(2\right)\left(4\right)=-15<0 \\ & \therefore \mbox{ no real roots}, \therefore \mbox{ no points of intersection} \end{align}$

d) $y=\frac{4}{x},y=5-x\,$

$\left(4,1\right) \mbox{ and }\left(1,4\right)$
- $\begin{align} & y=\frac{4}{x}\frac{\qquad}{}(1)\\ & y=5-x\frac{\qquad}{}(2)\\ & (1)\to(2):\\ & \frac{4}{x}=5-x \\ & 4=5x-x^{2} \\ & x^{2}-5x+4 =0 \\ &\left(x-4\right)\left(x-1\right)=0 \\ & x=4 , x=1 \\ & \mbox{When } x=4, y=1 ; x=1, y=4 \\ & \therefore \left(4,1\right) \mbox{ and }\left(1,4\right) \end{align}$

e) $y=\frac{3}{x}+1,y=x+3\,$

$\left(-3,0\right) \mbox{ and }\left(1,4\right)$
- $\begin{align} & y=\frac{3}{x}+1\frac{\qquad}{}(1)\\ & y=x+3\frac{\qquad}{}(2)\\ & (1)\to(2):\\ & \frac{3}{x}+1=x+3 \\ & 3+x=x^{2}+3x \\ & x^{2}+2x-3 =0 \\ &\left(x+3\right)\left(x-1\right)=0 \\ & x=-3 , x=1 \\ & \mbox{When } x=-3, y=0 ; x=1, y=4 \\ & \therefore \left(-3,0\right) \mbox{ and }\left(1,4\right) \end{align}$

f) $xy=1,y=2-x\,$

$\left(1,1\right)\mbox{ (touch)}$
- $\begin{align} & xy=1 \\ & y=\frac{1}{x}\frac{\qquad}{}(1)\\ & y=2-x\frac{\qquad}{}(2)\\ & (1)\to(2):\\ & \frac{1}{x}=2-x \\ & 1=2x-x^{2} \\ & x^{2}-2x+1 =0 \\ &\left(x-1\right)^{2}=0 \\ & x=1 \mbox{ (repeated)}, y=1 \\ & \therefore \left(1,1\right)\mbox{ (touch)} \end{align}$

g) $y=x^{3}-2x^{2}-x+3,y=4x-3\,$

$\left(1,1\right), \left(3,9\right) \mbox{ and }\left(-2,-11\right)$
- $\begin{align} & y=x^{3}-2x^{2}-x+3\frac{\qquad}{}(1)\\ & y=4x-3\frac{\qquad}{}(2)\\ & (1)\to(2):\\ & x^{3}-2x^{2}-x+3=4x-3 \\ & x^{3}-2x^{2}-5x+6=0 \\ &\left(x-1\right)\left(x^{2}-x-6\right)=0 \\ &\left(x-1\right)\left(x-3\right)\left(x+2\right)=0 \\ & x=1 , x=3 , x= -2 \\ & \mbox{When } x=1, y=1 ;x=3, y=9 ; x=-2, y=-11 \\ & \therefore \left(1,1\right), \left(3,9\right) \mbox{ and }\left(-2,-11\right) \end{align}$

h) $y=x^{3},3x-y-2=0\,$

$\therefore \left(1,1\right)\mbox{ (touch)} \mbox{ and }\left(-2,-8\right)$
- $\begin{align} & y=x^{3}\frac{\qquad}{}(1)\\ & 3x-y-2=0 \\ & y=3x-2\frac{\qquad}{}(2)\\ & (1)\to(2):\\ & x^{3}=3x-2 \\ & x^{3}-3x-2=0 \\ &\left(x-1\right)\left(x^{2}+x-2\right)=0 \\ &\left(x-1\right)\left(x-1\right)\left(x+2\right)=0 \\ &\left(x-1\right)^{2}\left(x+2\right)=0 \\ & x=1\mbox{ (repeated)} , x=-2\\ & \mbox{When } x=1, y=1; x=-2, y=-8 \\ & \therefore \left(1,1\right)\mbox{ (touch)} \mbox{ and }\left(-2,-8\right) \end{align}$

i) $y=x^{3}+6x^{2}-2x+10,y=2-14x\,$

$\left(-2,30\right)\mbox{ (tangent)}$
- $\begin{align} & y=x^{3}+6x^{2}-2x+10\frac{\qquad}{}(1)\\ & y=2-14x\frac{\qquad}{}(2)\\ & (1)\to(2):\\ & x^{3}+6x^{2}-2x+10=2-14x \\ & x^{3}+6x^{2}+12x+8=0 \\ &\left(x+2\right)\left(x^{2}+4x+4\right)=0 \\ &\left(x+2\right)\left(x+2\right)^{2}=0 \\ &\left(x+2\right)^{3}=0 \\ & x=-2 \mbox{ (thrice repeated)}, y=30\\ & \therefore \left(-2,30\right)\mbox{ (tangent)} \end{align}$

j) $y=x^{3}-x^{2},y=2x^{2}-x+3\,$

$\therefore \left(3,18\right)$
- $\begin{align} & y=x^{3}-x^{2}\frac{\qquad}{}(1)\\ & y=2x^{2}-x+3\frac{\qquad}{}(2)\\ & (1)\to(2):\\ & x^{3}-x^{2}=2x^{2}-x+3 \\ & x^{3}-3x^{2}+x-3=0 \\ &\left(x-3\right)\left(x^{2}+1\right)=0 \\ & x=3 , x^{2}+1=0 \mbox{ (rejected)}\\ & \mbox{When } x=3, y=18 \\ & \therefore \left(3,18\right) \end{align}$

k) $y=x,y=3-\frac{4}{x^{2}}\,$

$\left(-1,-1\right) \mbox{ and }\left(2,2\right)\mbox{ (touch)}$
- $\begin{align} & y=x\frac{\qquad}{}(1)\\ & y=3-\frac{4}{x^{2}}\frac{\qquad}{}(2)\\ & (1)\to(2):\\ & x=3-\frac{4}{x^{2}} \\ & x^{3}=3x^{2}-4 \\ & x^{3}-3x^{2}+4=0 \\ &\left(x+1\right)\left(x^{2}-4x+4\right)=0 \\ &\left(x+1\right)\left(x-2\right)^{2}=0 \\ & x=-1, x=2\mbox{ (repeated)}\\ & \mbox{When } x=-1, y=-1 ; x=2, y=2 \\ & \therefore \left(-1,-1\right) \mbox{ and }\left(2,2\right)\mbox{ (touch)} \end{align}$

l) $y=x-2,y=\frac{15}{x^{2}}-\frac{2}{x}\,$

$\left(3,11\right)$
- $\begin{align} & y=x-2\frac{\qquad}{}(1)\\ & y=\frac{15}{x^{2}}-\frac{2}{x}\frac{\qquad}{}(2)\\ & (1)\to(2):\\ & x-2=\frac{15}{x^{2}}-\frac{2}{x} \\ & x^{3}-2x^{2}+2x-15=0 \\ &\left(x-3\right)\left(x^{2}+x+5\right)=0 \\ &\mbox{For } x^{2}+x+5=0\\ & b^{2}-4ac = 1-4\left(5\right)=-19<0 \\ & \therefore \mbox{ no real roots}\\ & \therefore x=3, y=1 \\ & \therefore \left(3,11\right) \end{align}$

2) a) Prove that the line $y=2-x\,$ does not intersect the curve $y=\frac{15}{x}$ .

- $\begin{align} & y=2-x \frac{\qquad}{}(1) \\ & y=\frac{15}{x} \frac{\qquad}{}(2) \\ & (1)\to(2):\\ & 2-x=\frac{15}{x^{2}}\\ & 2x-x^{2}=15 \\ & x^{2}-2x+15 = 0\\ & b^{2}-4ac=\left(-2\right)^{2}-4\left(15\right)=-56<0 \\ & \therefore \mbox{ no real roots},\therefore \mbox{ they do not intersect}\\ \end{align}$

b) Prove that $y=x-3\,$ is tangent to the curve $y=-\frac{9}{4x}$ .

- $\begin{align} & y=x-3 \frac{\qquad}{}(1) \\ & y=-\frac{9}{4x} \frac{\qquad}{}(2) \\ & (1)\to(2):\\ & x-3=-\frac{9}{4x}\\ & 4x^{2}-12x=-9 \\ & 4x^{2}-12x+9 = 0\\ & \left(2x+3\right)^{2}=0 \\ & x=-\frac{2}{3}\mbox{ (repeated)}\\ & \mbox{ Since repeated roots},\therefore y=x-3 \mbox{ is tangent to the curve}\\ \end{align}$

c) Find the value of $k\,$ such that $y=k-x\,$ is a tangent to the curve $y=x^{2}\,$ .

$k=-\frac{1}{4}$
- $\begin{align} & y=k-x \frac{\qquad}{}(1) \\ & y=x^{2} \frac{\qquad}{}(2) \\ & (1)\to(2):\\ & k-x=x^{2}\\ & x^{2}+x-k=0 \\ & x^{2}-12x+9 = 0\\ & \mbox{tangent}, \therefore \mbox{ equal roots}\\ & \therefore b^{2}-4ac=0 \\ & 1-4\left(-k\right) =0 \\ & 4k=-1, k=-\frac{1}{4} \\ \end{align}$

d) Find the set of values of $p\,$ such that $y=5x-p\,$ intersect $y=4px^{2}\,$ at two different points.

$\left\{p:-\frac{5}{4}<p<0, 0<p<\frac{5}{4}\right\}$
- $\begin{align} & y=5x-p \frac{\qquad}{}(1) \\ & y=4px^{2} \frac{\qquad}{}(2) \\ & (1)\to(2):\\ & 5x-p=4px^{2}\\ & 4x^{2}-12x=-9 \\ & 4px^{2}-5x+p = 0\\ & \mbox{Intersect at two different points}, \therefore \mbox{two distinct real roots}\\ & \therefore b^{2}-4ac> 0 \qquad \mbox{ and } a \neq 0 \\ & 25 -4\left(4p\right)\left(p\right)>0 \qquad \quad p \neq 0 \\ & 25-16p^{2} >0 \\ & \left(5-4p\right)\left(5+4p\right)>0 \\ \end{align}$
- Media:MT-CG-E5-2d.dia
- $\begin{align} & \therefore -\frac{5}{4}<p<\frac{5}{4} \mbox{ and } p \neq 0 \\ & \therefore \left\{p:-\frac{5}{4}<p<0, 0<p<\frac{5}{4}\right\} \end{align}$

e) Find the value of $k\,$ such that the curve $y=x^{2}+x+2\,$ touches the curve $y=-x^{2}+2x-k\,$ .

$k=-\frac{15}{8}$
- $\begin{align} & y=x^{2}+x+2 \frac{\qquad}{}(1) \\ & y=-x^{2}+2x-k \frac{\qquad}{}(2) \\ & (1)\to(2):\\ & x^{2}+x+2=-x^{2}+2x-k\\ & 2x^{2}-x+k+2=0 \\ & \mbox{touches}, \therefore \mbox{ equal roots}\\ & \therefore b^{2}-4ac=0 \\ & 1-4\left(2\right)\left(k+2\right) =0 \\ & 1-8k-16=0 \\ & -15=8k \\ & k=-\frac{15}{8} \\ \end{align}$

f) Prove that the line $y=2k^{2}+3\,$ does not intersect the curve $y=-x^{2}+kx\,$ for all real values of $k\,$ .

- $\begin{align} & y=2k^{2}+3 \frac{\qquad}{}(1) \\ & y=-x^{2}+kx \frac{\qquad}{}(2) \\ & (1)\to(2):\\ & 2k^{2}+3=-x^{2}+kx\\ & x^{2}-kx+2k^{2}+3=0 \\ & b^{2}-4ac=k^{2}-4\left(2k^{2}+3\right) \\ & \qquad \quad\; = k^{2}-8k^{2}-12 \\ & \qquad \quad\; = -7k^{2}-12 \\ & \mbox{Since }k^{2} \ge 0 \mbox{ for all values of } k \\ & \therefore -7k^{2} \le 0 \\ & \therefore -7k^{2}-12 < 0 \\ & \therefore b^{2}-4ac < 0 \\ & \therefore \mbox{ no real roots} \\ & \therefore \mbox{ no intersection for all values of } k \\ \\ \end{align}$

Exercise 6

1. Find the radius & centre of the following circles

a) $x^{2}+y^{2}+2x-6y+2=0 \,$

$\mbox{centre is} \left(-1,3\right), \mbox{radius}=\sqrt{8}$
- $\begin{align} & x^{2}+y^{2}+2x-6y+2=0 \\ & \left(x+1\right)^{2}-1+\left(y-3\right)^{2}-9+2=0 \\ & \left(x+1\right)^{2}+\left(y-3\right)^{2}=8 \\ & \therefore \mbox{centre is} \left(-1,3\right), \mbox{radius}=\sqrt{8} \end{align}$
- OR
- $\begin{align} & 2g=2 \qquad 2f=-6 \\ & g=1 \qquad\; f=-3 \\ & \therefore \mbox{centre is} \left(-1,3\right)\\ & \mbox{radius}=\sqrt{1^{2}+\left(-3\right)^{2}-2}=\sqrt{8} \\ \end{align}$

b) $x^{2}+y^{2}+3x+y-1=0 \,$

$\mbox{centre is} \left(-\frac{3}{2},-\frac{1}{2}\right), \mbox{radius}=\sqrt{\frac{7}{2}}$
- $\begin{align} & x^{2}+y^{2}+3x+y-1=0 \\ & \left(x+\frac{3}{2}\right)^{2}-\frac{9}{4}+\left(y+\frac{1}{2}\right)^{2}-\frac{1}{4}-1=0 \\ & \left(x+\frac{3}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}=\frac{7}{2} \\ & \therefore \mbox{centre is} \left(-\frac{3}{2},-\frac{1}{2}\right), \mbox{radius}=\sqrt{\frac{7}{2}} \end{align}$
- OR
- $\begin{align} & 2g=3 \qquad 2f=1 \\ & g=\frac{3}{2} \qquad\; f=\frac{1}{2} \\ & \therefore \mbox{centre is} \left(-\frac{3}{2},-\frac{1}{2}\right)\\ & \mbox{radius}=\sqrt{\left(-\frac{3}{2}\right)^{2}+\left(-\frac{1}{2}\right)^{2}+1}=\sqrt{\frac{7}{2}} \\ \end{align}$

c) $3x^{2}+3y^{2}-4x-6y+3=0 \,$

$\mbox{centre is} \left(\frac{2}{3},1\right), \mbox{radius}=\frac{2}{3}$
- $\begin{align} & 3x^{2}+3y^{2}-4x-6y+3=0 \\ & x^{2}+y^{2}-\frac{4}{3}x-2y+1=0 \\ & \left(x-\frac{2}{3}\right)^{2}-\frac{4}{9}+\left(y-1\right)^{2}-1+1=0 \\ & \left(x-\frac{2}{3}\right)^{2}+\left(y-1\right)^{2}=\frac{4}{9} \\ & \therefore \mbox{centre is} \left(\frac{2}{3},1\right), \mbox{radius}=\sqrt{\frac{4}{9}}=\frac{2}{3} \end{align}$
- OR
- $\begin{align} & 2g=-\frac{4}{3} \qquad 2f=-2 \\ & g=-\frac{2}{3} \qquad\; f=-1 \\ & \therefore \mbox{centre is} \left(\frac{2}{3},1\right)\\ & \mbox{radius}=\sqrt{\left(-\frac{2}{3}\right)^{2}+\left(-1\right)^{2}-1}=\sqrt{\frac{4}{9}}=\frac{2}{3} \\ \end{align}$

2. Prove that the following circles have the stated centre and radius

a) $x^{2}+y^{2}-16x+4y+4=0 \,$ ; centre = $\left(8,-2\right)$ ; radius= $8\,$

- $\begin{align} & x^{2}+y^{2}-16x+4y+4=0 \\ & \left(x-8\right)^{2}-64+\left(y+2\right)^{2}-4+4=0 \\ & \left(x-8\right)^{2}+\left(y+2\right)^{2}=8^{2} \\ & \therefore \mbox{centre is} \left(8,-2\right), \mbox{radius}\frac{\qquad}{}\mbox{(Proved)} \end{align}$

b) $x^{2}+y^{2}-ax-by+c=0 \,$ ; centre = $\left(\frac{a}{2},\frac{b}{2}\right)$ ; radius= $\sqrt{\frac{a^{2}}{4}+\frac{b^{2}}{4}-c}$

- $\begin{align} & x^{2}+y^{2}-ax-by+c=0 \\ & \left(x-\frac{a}{2}\right)^{2}-\frac{a^{2}}{4}+\left(y-\frac{b}{2}\right)^{2}-\frac{b^{2}}{4}+c=0 \\ & \left(x-\frac{a}{2}\right)^{2}+\left(y-\frac{b}{2}\right)^{2}=\left(\sqrt{\frac{a^{2}}{4}+\frac{b^{2}}{4}-c}\right)^{2} \\ & \therefore \mbox{centre is} \left(\frac{a}{2},\frac{b}{2}\right), \mbox{radius}=\sqrt{\frac{a^{2}}{4}+\frac{b^{2}}{4}-c}\frac{\qquad}{}\mbox{(Proved)} \end{align}$

3. Find the equation of the circle with the following properties:

a) centre $\left(-3,2\right)$ and radius $5\,$

$x^{2}+y^{2}+6x-4y-12=0\,$
- $\begin{align} & \left(x+3\right)^{2}+\left(y-2\right)^{2}=5^{2}\\ & x^{2}+6x+9+y^{2}-4y+4=25 \\ & x^{2}+y^{2}+6x-4y-12=0 \\ \end{align}$

b) centre $\left(4,-1\right)$ and radius $2\sqrt{7}\,$

$x^{2}+y^{2}-8x+2y-11=0\,$
- $\begin{align} & \left(x-4\right)^{2}+\left(y+1\right)^{2}=\left(2\sqrt{7}\right)^{2}\\ & x^{2}-8x+16+y^{2}+2y+1=4\left(7\right) \\ & x^{2}+y^{2}-8x+2y-11=0 \\ \end{align}$

c) diameter $AB\,$ . $A\left(1,3\right)$ and $B\left(-2,0\right)$

$x^{2}+y^{2}+x-3y-2=0\,$
- File:MT-CG-E6-3c.dia
- $\begin{align} & \mbox{Let centre} = C \\ & \therefore C = \mbox{Midpoint of } AB \\ & \qquad = \left(\frac{1-2}{2}, \frac{3}{2}\right)\\ & \qquad = \left(-\frac{1}{2}, \frac{3}{2}\right)\\ & r = \frac{1}{2}AB \\ & \quad = \frac{1}{2}\sqrt{\left(1+2\right)^{2}+3^{2}} = \frac{1}{2}\sqrt{18}\\ & \therefore \left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}=\left(\frac{1}{2}\sqrt{18}\right)^{2}\\ & x^{2}+x+\frac{1}{4}+y^{2}-3y+\frac{9}{4}=\frac{18}{4} \\ & x^{2}+y^{2}+x-3y-2=0 \\ \end{align}$

d) centre $\left(2,1\right)$ and touching the line $5x-12y+10=0\,$

$x^{2}+y^{2}-4x-2y+\frac{781}{169}=0 \,$
- File:MT-CG-E6-3d.dia
- $\begin{align} & r =d \\ & \;\; = \frac{\left|5\left(2\right)-12\left(1\right)+10\right|}{\sqrt{5^{2}+\left(-12\right)^{2}}}= \frac{8}{13}\\ & \therefore \left(x-2\right)^{2}+\left(y-1\right)^{2}=\left(\frac{8}{13}\right)^{2}\\ & x^{2}-2x+4+y^{2}-2y+1=\frac{64}{169} \\ & x^{2}+y^{2}-4x-2y+\frac{781}{169}=0 \\ \end{align}$

e) centre $\left(-3,0\right)$ and tangential to the line $y=4x-3\,$

$x^{2}+y^{2}+6x-\frac{72}{17}=0 \,$
- File:MT-CG-E6-3d.dia
- $\begin{align} & y=4x-3, \therefore 4x-y-3=0 \\ & r =d \\ & \;\; = \frac{\left|4\left(-3\right)-1\left(0\right)-3\right|}{\sqrt{4^{2}+\left(-1\right)^{2}}}= \frac{15}{\sqrt{17}}\\ & \therefore \left(x+3\right)^{2}+y^{2}=\left(\frac{15}{\sqrt{17}}\right)^{2}\\ & x^{2}+6x+9+y^{2}=\frac{225}{17} \\ & x^{2}+y^{2}+6x-\frac{72}{17}=0 \\ \end{align}$

f) radius $5\,$ and touching both axis

$x^{2}+y^{2}-10x-10y+25=0;x^{2}+y^{2}-10x+10y+25=0;\,$ $x^{2}+y^{2}+10x-10y+25=0;x^{2}+y^{2}+10x+10y+25=0\,$
- File:MT-CG-E6-3f.dia
- $\begin{align} & \therefore r=5, \mbox{Centre }= \left(5,5\right),\left(-5,5\right),\left(5,-5\right) \mbox{ or } \left(-5,-5\right)\\ & \left(5,5\right):\left(x-5\right)^{2}+\left(y-5\right)^{2}=5^{2}\\ & \therefore x^{2}+y^{2}-10x-10y+25=0 \\ & \left(5,-5\right):\left(x-5\right)^{2}+\left(y+5\right)^{2}=5^{2}\\ & \therefore x^{2}+y^{2}-10x+10y+25=0 \\ & \left(-5,5\right):\left(x+5\right)^{2}+\left(y-5\right)^{2}=5^{2}\\ & \therefore x^{2}+y^{2}+10x-10y+25=0 \\ & \left(-5,-5\right):\left(x+5\right)^{2}+\left(y+5\right)^{2}=5^{2}\\ & \therefore x^{2}+y^{2}+10x+10y+25=0 \\ \end{align}$

g) passing through $\left(2,0\right), \left(0,3\right)$ and $\left(5,2\right)$

$x^{2}+y^{2}-5x-5y+6=0 \,$
- $\begin{align} & \mbox{Let equation}: x^{2}+y^{2}+ax+by+c=0 \\ & \left(2,0\right) : 4 +2a +c=0 \\ & \therefore c=-4-2a \frac{\qquad}{}(1) \\ & \left(0,3\right) : 9+3b+c =0\frac{\qquad}{}(2) \\ & \left(5,2\right) : 29+5a+2b+c=0\frac{\qquad}{}(3) \\ & (1) \to (2) : 9+3b-4-2a =0 \\ & \qquad \qquad \quad 3b-2a=-5\frac{\qquad}{}(4)\\ & (1) \to (3) : 29+5a+2b-4-2a=0 \\ & \qquad \qquad \quad 3a+2b=-25\frac{\qquad}{}(5) \\ & (4)\times 2 : 6b-4a=-10\frac{\qquad}{}(6) \\ & (5)\times 3 : 9a+6b=-75\frac{\qquad}{}(7) \\ & (7)-(6): 13a=-65, \therefore a =-5\\ & \therefore b=-5, c=6 \\ & \therefore x^{2}+y^{2}-5x-5y+6=0 \\ \end{align}$

h) passing through $\left(4,3\right), \left(9,-2\right)$ and $\left(8,1\right)$

$x^{2}+y^{2}-8x+4y-5=0 \,$
- $\begin{align} & \mbox{Let equation}: x^{2}+y^{2}+ax+by+c=0 \\ & \left(4,3\right) : 25+4a+3b+c=0 \\ & \therefore c=-25-4a-3b \frac{\qquad}{}(1) \\ & \left(9,-2\right) : 85+9a-2b+c=0\frac{\qquad}{}(2) \\ & \left(8,1\right) : 65+8a+b+c=0\frac{\qquad}{}(3) \\ & (1) \to (2) : 85+9a-2b-25-4a-3b =0 \\ & \qquad \qquad \quad 60+5a-5b=0 \\ & \qquad \qquad \quad 12+a-b=0\frac{\qquad}{}(4)\\ & (1) \to (3) : 65+8a+b-25-4a-3b=0 \\ & \qquad \qquad \quad 40+4a-2b=0\\ & \qquad \qquad \quad 20+2a-b=0\frac{\qquad}{}(5) \\ & (5)-(4): 8+a=0, \therefore a =-8\\ & \therefore b=4, c=-5 \\ & \therefore x^{2}+y^{2}-8x+4y-5=0 \\ \end{align}$

${\color{Red}****}$ 4. Find the equation of the tangent and normal to the following circle at the following points on the circle

a) $x^{2}+y^{2}-6x+4y+3=0;\left(0,-1\right)$

$y=3x-1; y=-\frac{1}{3}x-1$
- Media:MT-CG-E6-4a.dia
- $\begin{align} & \mbox{Let } A\left(0,-1\right) \\ & x^{2}+y^{2}-6x+4y+3=0 \\ & \left(x-3\right)^{2}+\left(y+2\right)^{2}=10 \\ & \mbox{Centre } =\left(3,-2\right)\\ & m_{AC} =\frac{-2+1}{3-0}=-\frac{1}{3}\\ & \mbox{Let gradient of tangent } = m_{1} \\ & m_{1}m_{AC}=-1 \\ & m_{1} = 3 \\ & \mbox{equation of tangent}: y+1=3\left(x-0\right) \\ & y=3x-1\\ & \mbox{equation of normal}: y+1=-\frac{1}{3}\left(x-0\right) \\ & y=-\frac{1}{3}x-1\\ \end{align}$

b) $x^{2}+y^{2}+8x+6y+5=0;\left(-6,1\right)$

$y=\frac{1}{2}x+4; y=-2x-11$
- Media:MT-CG-E6-4a.dia
- $\begin{align} & \mbox{Let } A\left(-6,1\right) \\ & x^{2}+y^{2}+8x+6y+5=0 \\ & \left(x+4\right)^{2}+\left(y+3\right)^{2}=20 \\ & \mbox{Centre } =\left(-4,-3\right)\\ & m_{AC} =\frac{-3-1}{-4+6}=-2\\ & \mbox{Let gradient of tangent } = m_{1} \\ & m_{1}m_{AC}=-1 \\ & m_{1} = -\frac{1}{2} \\ & \mbox{equation of tangent}: y-1=\frac{1}{2}\left(x+6\right) \\ & y=\frac{1}{2}x+4\\ & \mbox{equation of normal}: y-1=-2\left(x+6\right) \\ & y=-2x-11\\ \end{align}$

c) $x^{2}+y^{2}-6x+3y-5=0;\left(-1,-2\right)$

$y=-8x-10; y=\frac{1}{8}x-\frac{15}{8}$
- Media:MT-CG-E6-4a.dia
- $\begin{align} & \mbox{Let } A\left(-1,-2\right) \\ & x^{2}+y^{2}-6x+3y-5=0 \\ & \left(x-3\right)^{2}+\left(y+\frac{3}{2}\right)^{2}=\frac{65}{4} \\ & \mbox{Centre } =\left(3,-\frac{3}{2}\right)\\ & m_{AC} =\frac{-2+\frac{3}{2}}{-1-3}=\frac{1}{8}\\ & \mbox{Let gradient of tangent } = m_{1} \\ & m_{1}m_{AC}=-1 \\ & m_{1} = -8 \\ & \mbox{equation of tangent}: y+2=-8\left(x+1\right) \\ & y=-8x-10\\ & \mbox{equation of normal}: y+2=\frac{1}{8}\left(x+1\right) \\ & y=\frac{1}{8}x-\frac{15}{8}\\ \end{align}$

5. Find the equation of the tangent from the origin to the following circle

a) $x^{2}+y^{2}-4x-8y+10=0\,$

$y=\frac{1}{3}x \mbox{ and }y=3x\,$
- $\begin{align} & \mbox{Let tangent be } y=mx\frac{\qquad}{}(1) \\ & x^{2}+y^{2}-4x-8y+10=0 \frac{\qquad}{}(2)\\ & (1) \to (2) : x^{2}+\left(mx\right)^{2}-4x-8\left(mx\right)+10=0 \\ & x^{2}+m^{2}x^{2}-4x-8mx+10=0\\ & \left(m^{2}+1\right)x^{2}+4\left(2m+1\right)+10=0\\ & \mbox{Since tangent,} \therefore \mbox{equal roots}\\ & \therefore b^{2}-4ac=0\\ & \left[4\left(2m+1\right)\right]^{2}-4\left(m^{2}+1\right)\left(10\right)=0\\ & 16\left(4m^{2}+4m+1\right)-40\left(m^{2}+1\right)=0\\ & 8m^{2}+8m+2-5m^{2}-5=0\\ & 3m^{2}+8m-3 =0 \\ & \left(3m-1\right)\left(m+3\right)=0\\ & m=\frac{1}{3} \mbox{ or } m=-3\\ & \therefore y=\frac{1}{3}x \mbox{ and }y=3x \end{align}$

b) $x^{2}+y^{2}+6x-10y+17=0\,$

$y=-\frac{1}{4}x \mbox{ and }y=4x\,$
- $\begin{align} & \mbox{Let tangent be } y=mx\frac{\qquad}{}(1) \\ & x^{2}+y^{2}+6x-10y+17=0 \frac{\qquad}{}(2)\\ & (1) \to (2) : x^{2}+\left(mx\right)^{2}+6x-10\left(mx\right)+17=0 \\ & x^{2}+m^{2}x^{2}+6x-10mx+17=0\\ & \left(m^{2}+1\right)x^{2}-2\left(5m-3\right)+17=0\\ & \mbox{Since tangent,} \therefore \mbox{equal roots}\\ & \therefore b^{2}-4ac=0\\ & \left[-2\left(5m-3\right)\right]^{2}-4\left(m^{2}+1\right)\left(17\right)=0\\ & \left(25m^{2}-30m+9\right)-17\left(m^{2}+1\right)=0\\ & 25m^{2}-30m+9-17m^{2}-17=0\\ & 8m^{2}-30m-8 =0 \\ & 4m^{2}-15m-4 =0 \\ & \left(4m+1\right)\left(m-4\right)=0\\ & m=-\frac{1}{4} \mbox{ or } m=4\\ & \therefore y=-\frac{1}{4}x \mbox{ and }y=4x \end{align}$

6. a) Find equation of tangents to the circle $x^{2}+y^{2}-4x-4y-2=0\,$ which have gradient $3\,$

$y=3x+6 \mbox{ and }y=3x-14\,$
- $\begin{align} & \mbox{Let tangent be } y=3x+c\frac{\qquad}{}(1) \\ & x^{2}+y^{2}-4x-4y-2=0 \frac{\qquad}{}(2)\\ & (1) \to (2) : x^{2}+\left(3x+c\right)^{2}-4x-4\left(3x+c\right)-2=0 \\ & x^{2}+9x^{2}+6cx+c^{2}-4x-12x-4c-2=0\\ & 10x^{2}+2\left(3c-8\right)x+\left(c^{2}-4c-2\right)=0\\ & \mbox{Since tangent,} \therefore \mbox{equal roots}\\ & \therefore b^{2}-4ac=0\\ & \left[2\left(3c-8\right)\right]^{2}-4\left(10\right)\left(c^{2}-4c-2\right)=0\\ & \left(9c^{2}-48c+64\right)-10\left(c^{2}-4c-2\right)=0\\ & 9c^{2}-48c+64-10c^{2}+40c+20\\ & -c^{2}-8c+84 =0 \\ & c^{2}+8c-84 =0 \\ & \left(c-6\right)\left(c+14\right)=0\\ & c=6 \mbox{ or } c=-14\\ & \therefore y=3x+6 \mbox{ and }y=3x-14 \end{align}$

b) Find the equation of tangents to the circle $x^{2}+y^{2}=289\,$ that are parallel to the line $8x=15y\,$

$y=3x+6 \mbox{ and }y=3x-14\,$
- $\begin{align} & 8x=15y, m=\frac{8}{15}\\ & \therefore m_{tangent}=\frac{8}{15}\\ & \mbox{Let tangent be } y=\frac{8}{15}x+c\frac{\qquad}{}(1) \\ & x^{2}+y^{2}=289 \frac{\qquad}{}(2)\\ & (1) \to (2) : x^{2}+\left(\frac{8}{15}x+c\right)^{2}=289 \\ & x^{2}+\frac{64}{225}x^{2}+\frac{16}{15}cx+c^{2}=289\\ & \frac{289}{225}x^{2}+\frac{16}{15}cx+c^{2}-289=0\\ & \mbox{Since tangent,} \therefore \mbox{equal roots}\\ & \therefore b^{2}-4ac=0\\ & \left(\frac{16}{15}c\right)^{2}-4\left(\frac{289}{225}\right)\left(c^{2}-289\right)=0\\ & 256c^{2}-4\left(289\right)\left(c^{2}-289\right)=0\\ & 64c^{2}-289c^{2}+289^{2}=0 \\ & 225c^{2}=289^{2}\\ & c=\pm\frac{289}{15} \\ & c=\frac{289}{15} \mbox{ or } c=-\frac{289}{15}\\ & \therefore y=\frac{8}{15}x+\frac{289}{15} \mbox{ and }y=\frac{8}{15}x-\frac{289}{15} \end{align}$

7. a) Prove that $3x-4y-3=0\,$ is tangent to the circle $x^{2}+y^{2}-4x+6y+4=0\,$ . Find its point of contact.

$\left(\frac{1}{5},-\frac{3}{5}\right)$
- Media:MT-CG-E6-3d.dia
- $\begin{align} & x^{2}+y^{2}-4x+6y+4=0 \\ & \left(x-2\right)^{2}+\left(y+3\right)^{2}=9 \\ & \mbox{Centre } = \left(2,-3\right), r=3 \\ & \left(2,3\right), 3x-4y-3=0:\\ & d=\frac{\left|3\left(2\right)-4\left(-3\right)-3\right|}{\sqrt{3^{2}+\left(-4\right)^{2}}}=3 \\ & \mbox{Since } r=d, \therefore 3x-4y-3=0\mbox{ is tangent to the circle}\\ & 3x-4y-3=0 \\ & \therefore y=\frac{3}{4}\left(x-1\right)\frac{\qquad}{}(1)\\ & x^{2}+y^{2}-4x+6y+4=0\frac{\qquad}{}(2)\\ & (2)\to(1):\\ & x^{2}+\left[\frac{3}{4}\left(x-1\right)\right]^{2}-4x+6\left[\frac{3}{4}\left(x-1\right)\right]+4=0 \\ & x^{2}+\frac{9}{16}\left(x^{2}-2x+1\right)-4x+\frac{9}{2}\left(x-1\right)+4=0\\ & 16x^{2}+9x^{2}-18x+9-64x+72x-72+64=0\\ & 25x^{2}-10x+1 \\ & \left(5x-1\right)^{2} =0\\ & \therefore x=\frac{1}{5}, y =-\frac{3}{5} \\ & \mbox{Point of contact } = \left(\frac{1}{5},-\frac{3}{5}\right) \end{align}$

b) Determine whether $y=4x-15\,$ is tangent to the circle $x^{2}+y^{2}=16\,$ .

- $\begin{align} & x^{2}+y^{2}=16 \\ & \mbox{Centre } = \left(2,-3\right), r=3 \\ & \left(0,0\right), 4x-y-15=0:\\ & d=\frac{\left|4\left(0\right)-1\left(0\right)-15\right|}{\sqrt{4^{2}+\left(-1\right)^{2}}}=\frac{15}{\sqrt{17}} \\ & \mbox{Since } r \neq d, \therefore y=4x-15 \mbox{ is not tangent to the circle}\\ \end{align}$

8. Find the length of the tangent

a) from $\left(0,2\right)$ to the circle $x^{2}+y^{2}+12x-10y+56=0\,$

$2\sqrt{10}$
- File:MT-CG-E6-8a.dia
- $\begin{align} & x^{2}+y^{2}+12x-10y+56=0\\ & \left(x+6\right)^{2}+\left(y-5\right)^{2}=5 \\ & \therefore C\left(-6,5\right), r=\sqrt{5} , A\left(0,2\right)\\ & AC =\sqrt{6^{2}+3^{2}}=\sqrt{45}\\ & AC^{2}=r^{2}+l^{2}\\ & 45 = 5 +l^{2} \\ & l =\sqrt{40}=2\sqrt{10} \\ \end{align}$

b) from $\left(5,-2\right)$ to the circle $x^{2}+y^{2}-2x-4y+1=0\,$

$2\sqrt{7}$
- File:MT-CG-E6-8a.dia
- $\begin{align} & x^{2}+y^{2}-2x-4y+1=0\\ & \left(x-1\right)^{2}+\left(y-2\right)^{2}=4 \\ & \therefore C\left(1,2\right), r=2 , A\left(5,-2\right)\\ & AC =\sqrt{4^{2}+4^{2}}=\sqrt{32}\\ & AC^{2}=r^{2}+l^{2}\\ & 32 = 4 +l^{2} \\ & l =\sqrt{28}=2\sqrt{7} \\ \end{align}$

9. Find the equation of the circle(s) with the following properties

a) centre lies on the line $y=3x-2\,$ radius is $\sqrt{17}$ and passes through $\left(-2,5\right)$ .

$x^{2}+y^{2}-4x-8y+3 =0 \mbox{ and }5x^{2}+5y^{2}-18x-34y-11 =0\,$
- Media:MT-CG-E6-9a.dia
- $\begin{align} & \mbox{Centre lies on } y=3x-2 \\ & \mbox{When }x=a, y=3a-2 \\ & \therefore \mbox{Let } C\left(a,3a-2\right), A\left(-2,5\right)\\ & AC=r \\ & \sqrt{\left(a+2\right)^{2}+\left(3a-2-5\right)^{2}}=\sqrt{17} \\ & a^{2}+4a+4+9a^{2}-42a+49=17 \\ & 10a^{2}-38x+36 = 0 \\ & 5a^{2}-19x+18 = 0\\ & \left(a-2\right)\left(5a-9\right)=0\\ & a=2 \mbox{ or } a=\frac{9}{5} \\ & \mbox{When } a=2, C\left(2,4\right), r=\sqrt{17}:\\ & \left(x-2\right)^{2}+\left(y-4\right)^{2}=17 \\ & \therefore x^{2}+y^{2}-4x-8y+3 =0 \\ & \mbox{When } a=\frac{9}{5}, C\left(\frac{9}{5},\frac{17}{5}\right), r=\sqrt{17}:\\ & \left(x-\frac{9}{5}\right)^{2}+\left(y-\frac{17}{5}\right)^{2}=17 \\ & x^{2}+y^{2}-\frac{18}{5}x-\frac{34}{5}y-\frac{11}{5} =0 \\ & \therefore 5x^{2}+5y^{2}-18x-34y-11 =0 \\ & \therefore x^{2}+y^{2}-4x-8y+3 =0 \mbox{ and }5x^{2}+5y^{2}-18x-34y-11 =0 \\ \end{align}$

b) centre lies on the line $x-y-4=0\,$ and it is tangent to the line $2x-3y+4=0\,$ at $\left(1,2\right)$

$x^{2}+y^{2}-6x+2y-3 =0 \,$
- Media:MT-CG-E6-9b.dia
- $\begin{align} & \mbox{Centre lies on } x-y-4=0 \\ & \mbox{When }x=a, y=a-4 \\ & \therefore \mbox{Let } C \left(a,a-4\right), A\left(1,2\right)\\ & 2x-3y+4=0, m_{1}=\frac{2}{3}\\ & m_{AC}m_{1}=-1\\ & \therefore m_{AC}=-\frac{3}{2}\\ & \frac{\left(a-4\right)-2}{a-1}=-\frac{3}{2}\\ & \frac{a-6}{a-1}=-\frac{3}{2} \\ & 2a-12=3-3a \\ & 5a=15, \therefore a=3 \\ & \therefore C\left(3,-1\right) r=AC=\sqrt{2^{2}+3^{2}}=\sqrt{13}\\ & \left(x-3\right)^{2}+\left(y+1\right)^{2}=13 \\ & \therefore x^{2}+y^{2}-6x+2y-3 =0 \\ \end{align}$

c) Touching both axis and tangential to the line $5x+12y=60\,$

$x^{2}+y^{2}-30x-30y+225 =0 \mbox{ and }x^{2}+y^{2}-4x-4y+4 =0\,$
- Media:MT-CG-E6-9c.dia
- $\begin{align} & \therefore \mbox{Let } C \left(r,r\right)\\ & r=d \\ & \left(r,r\right), 5x+12y-60=0:\\ & r =\frac{\left|5r+12r-60\right|}{5^{2}+12^{2}}\\ & \quad = \frac{\left|17r-60\right|}{13}\\ & 13r = \left|17r-60\right| \\ & 13r = 17r-60 \mbox{ or } 13r = -17r+60 \\ & 4r=60 \qquad \qquad \quad 30r=60 \\ & r=15 \qquad \qquad \qquad r=2 \\ & \mbox{When } r=15, C\left(15,15\right)\\ & \left(x-15\right)^{2}+\left(y-15\right)^{2}=15^{2} \\ & \therefore x^{2}+y^{2}-30x-30y+225 =0 \\ & \mbox{When } r=2, C\left(2,2\right)\\ & \left(x-2\right)^{2}+\left(y-2\right)^{2}=2^{2} \\ & \therefore x^{2}+y^{2}-4x-4y+4 =0 \\ & \therefore x^{2}+y^{2}-30x-30y+225 =0 \mbox{ and }x^{2}+y^{2}-4x-4y+4 =0 \\ \end{align}$

Exercise 7

1. Sketch the graph of the following circles

a) $x^{2}+y^{2}=4\,$

- File:MT-CG-E7-1a.dia
- $\mbox{Centre }=\left(0,0\right), r=2$

b) $x^{2}+y^{2}=3\,$

- File:MT-CG-E7-1b.dia
- $\mbox{Centre }=\left(0,0\right), r=\sqrt{3}$

c) $\left(x-3\right)^{2}+\left(y+2\right)^{2}=16\,$

- File:MT-CG-E7-1c.dia
- $\mbox{Centre }=\left(3,-2\right), r=4$

d) $x^{2}+y^{2}-2x-1=0\,$

- File:MT-CG-E7-1d.dia
- $\begin{align} & x^{2}+y^{2}-2x-1=0 \\ & \left(x-1\right)^{2}+y^{2}=2 \\ & \mbox{Centre }=\left(1,0\right), r=\sqrt{2} \end{align}$

e) $x^{2}+y^{2}-3x-6y+5=0\,$

- File:MT-CG-E7-1e.dia
- $\begin{align} & x^{2}+y^{2}-3x-6y+5=0 \\ & \left(x-\frac{3}{2}\right)^{2}+\left(y-3\right)=\frac{25}{4} \\ & \mbox{Centre }=\left(\frac{3}{2},3\right), r=\frac{5}{2} \end{align}$

2. State the centre and length of the major and minor axis of the following ellipse. Sketch the graph of each ellipse.

a) $\frac{x^{2}}{4}+\frac{y^{2}}{25}=1$

- $\begin{align} & \frac{x^{2}}{4}+\frac{y^{2}}{25}=1 \\ & \mbox{Centre }=\left(0,0\right)\\ & \mbox{length of major axis }=10\\ & \mbox{length of minor axis }=4\\ \end{align}$
- File:MT-CG-E7-2a.dia

b) $16x^{2}+8y^{2}=1\,$

- $\begin{align} & \mbox{Centre }=\left(0,0\right)\\ & \mbox{length of major axis }=\frac{1}{2}\sqrt{2}\\ & \mbox{length of minor axis }=\frac{1}{2}\\ \end{align}$
- File:MT-CG-E7-2b.dia
- $\begin{align} & 16x^{2}+8y^{2}=1 \\ & \frac{x^{2}}{\left(\frac{1}{16}\right)}+\frac{y^{2}}{\left(\frac{1}{8}\right)}=1 \\ & a=\frac{1}{4}, b=\frac{1}{\sqrt{8}}=\frac{\sqrt{8}}{8}=\frac{1}{4}\sqrt{2}\\ & \mbox{Centre }=\left(0,0\right)\\ & \mbox{length of major axis }=\frac{1}{2}\sqrt{2}\\ & \mbox{length of minor axis }=\frac{1}{2}\\ \end{align}$

c) $x^{2}+9y^{2}=4\,$

- $\begin{align} & \mbox{Centre }=\left(0,0\right)\\ & \mbox{length of major axis }=4\\ & \mbox{length of minor axis }=\frac{4}{3}\\ \end{align}$
- File:MT-CG-E7-2c.dia
- $\begin{align} & x^{2}+9y^{2}=4\\ & \frac{x^{2}}{4}+\frac{9y^{2}}{4}=1\\ & \frac{x^{2}}{4}+\frac{y^{2}}{\left(\frac{4}{9}\right)}=1\\ & a=2, b=\frac{2}{3}\\ & \mbox{Centre }=\left(0,0\right)\\ & \mbox{length of major axis }=4\\ & \mbox{length of minor axis }=\frac{4}{3}\\ \end{align}$

d) $\frac{\left(x+3\right)^{2}}{4}+\left(y-1\right)^{2}=1\,$

- $\begin{align} & \mbox{Centre }=\left(-3,1\right)\\ & \mbox{length of major axis }=4\\ & \mbox{length of minor axis }=2\\ \end{align}$
- File:MT-CG-E7-2d.dia

e) $\frac{\left(x+1\right)^{2}}{8}+\frac{4y^{2}}{25}=1\,$

- $\begin{align} & \mbox{Centre }=\left(-1,0\right)\\ & \mbox{length of major axis }=4\sqrt{2}\\ & \mbox{length of minor axis }=5\\ \end{align}$
- File:MT-CG-E7-2e.dia
- $\begin{align} & \frac{\left(x+1\right)^{2}}{8}+\frac{4y^{2}}{25}=1\\ & \frac{\left(x+1\right)^{2}}{8}+\frac{y^{2}}{\left(\frac{25}{4}\right)}=1\\ & a=\sqrt{8}=2\sqrt{2}, b=\frac{5}{2}\\ & \mbox{Centre }=\left(-1,0\right)\\ & \mbox{length of major axis }=4\sqrt{2}\\ & \mbox{length of minor axis }=5\\ \end{align}$

f) $9x^{2}+4y^{2}-18x+16y-11=0\,$

- $\begin{align} & \mbox{Centre }=\left(1,-2\right)\\ & \mbox{length of major axis }=6\\ & \mbox{length of minor axis }=4\\ \end{align}$
- File:MT-CG-E7-2f.dia
- $\begin{align} & 9x^{2}+4y^{2}-18x+16y-11=0 \\ & 9\left(x^{2}-2x\right)+4\left(y^{2}+4y\right)-11=0\\ & 9\left[\left(x-1\right)^{2}-1\right]+4\left[\left(y+2\right)^{2}-4\right]-11=0\\ & 9\left(x-1\right)^{2}+4\left(y+2\right)^{2}=36\\ & \frac{\left(x-1\right)^{2}}{4}+\frac{\left(y+2\right)^{2}}{9}=1\\ & a=2, b=3\\ & \mbox{Centre }=\left(1,-2\right)\\ & \mbox{length of major axis }=6\\ & \mbox{length of minor axis }=4\\ \end{align}$

3. Find the tangent with the following properties to the following ellipses. In each case, determine the point of contact.

a) with gradient $2\,$ . Ellipse : $3x^{2}+5y^{2}=1\,$ .

$\begin{align} & y=2x+\frac{\sqrt{345}}{15} \mbox{ and }y=2x-\frac{\sqrt{345}}{15}\\ & \left(-\frac{2}{69}\sqrt{345},\frac{\sqrt{345}}{115}\right),\left(\frac{2}{69}\sqrt{345},-\frac{\sqrt{345}}{115}\right) \end{align}$
- $\begin{align} & \mbox{Let tangent be } y=2x+c\frac{\qquad}{}(1) \\ & 3x^{2}+5y^{2}=1=0 \frac{\qquad}{}(2)\\ & (1) \to (2) : 3x^{2}+5\left(2x+c\right)^{2}=1 \\ & 3x^{2}+5\left(4x^{2}+4cx+c^{2}\right)=1 \\ & 3x^{2}+20x^{2}+20cx+5c^{2}=1 \\ & 23x^{2}+20cx+5c^{2}-1=0 \\ & \mbox{Since tangent,} \therefore \mbox{equal roots}\\ & \therefore b^{2}-4ac=0\\ & 400c^{2}-4\left(23\right)\left(5c^{2}-1\right)=0\\ & 100c^{2}-115c^{2}+23 =0\\ & -15c^{2}=23 \\ & c=\pm \sqrt{\frac{23}{15}}=\frac{\sqrt{345}}{15}\\ & \therefore y=2x+\frac{\sqrt{345}}{15} \mbox{ and }y=2x-\frac{\sqrt{345}}{15}\\ \end{align}$
- $\begin{align} & \mbox{When } c=\frac{\sqrt{345}}{15} :\\ & 23x^{2}+20\left(\frac{\sqrt{345}}{15}\right)x+5\left(\frac{23}{15}\right)-1=0 \\ & 23x^{2}+\frac{4}{3}\sqrt{345}x+\frac{20}{3}=0 \\ & 69x^{2}+4\sqrt{345}x+20=0 \\ & x=\frac{-4\sqrt{345}\pm\sqrt{\left(4\sqrt{345}\right)^{2}-4\left(69\right)\left(20\right)}}{138}\\ & \quad =-\frac{2}{69}\sqrt{345}\\ & y=2\left(-\frac{2}{69}\sqrt{345}\right)+\frac{\sqrt{345}}{15}=\frac{\sqrt{345}}{115}\\ & \mbox{When } c=-\frac{\sqrt{345}}{15} :\\ & 23x^{2}+20\left(-\frac{\sqrt{345}}{15}\right)x+5\left(\frac{23}{15}\right)-1=0 \\ & 69x^{2}-4\sqrt{345}x+20=0 \\ & x=\frac{4\sqrt{345}\pm\sqrt{\left(-4\sqrt{345}\right)^{2}-4\left(69\right)\left(20\right)}}{138}\\ & \quad =\frac{2}{69}\sqrt{345}\\ & y=2\left(\frac{2}{69}\sqrt{345}\right)-\frac{\sqrt{345}}{15}=-\frac{\sqrt{345}}{115}\\ & \therefore \mbox{points of contact are } \left(-\frac{2}{69}\sqrt{345},\frac{\sqrt{345}}{115}\right),\left(\frac{2}{69}\sqrt{345},-\frac{\sqrt{345}}{115}\right) \end{align}$

b) passes through $\left(3,0 \right)$ . $x^{2}+2y^{2}=1\,$ .

$\begin{align} & y=\frac{1}{4}x-\frac{3}{4} \mbox{ and }y=-\frac{1}{4}x+\frac{3}{4}\\ & \left(\frac{1}{3},-\frac{2}{3}\right), \left(\frac{1}{3},\frac{2}{3}\right)\\ \end{align}$
- $\begin{align} & \mbox{Let gradient of tangent } =m \\ & \left(3,0 \right) : y-0=m\left(x-3\right)\\ & y=mx-3m\frac{\qquad}{}(1)\\ & x^{2}+2y^{2}=1 \frac{\qquad}{}(2)\\ & (1) \to (2) : x^{2}+2\left(mx-3m\right)^{2}=1 \\ & x^{2}+2\left(m^{2}x^{2}-6mx+9m^{2}\right)-1=0\\ & \left(2m^{2}+1\right)x^{2}-12m^{2}x+18m^{2}-1=0\\ & \mbox{Since tangent,} \therefore \mbox{equal roots}\\ & \therefore b^{2}-4ac=0\\ & \left(-12m^{2}\right)^{2}-4\left(2m^{2}+1\right)\left(18m^{2}-1\right)=0\\ & 144m^{4}-4\left(36m^{4}+16m^{2}-1\right)=0\\ & 16m^{2}-1=0\\ & m^{2} =\frac{1}{16}, \therefore m =\pm\frac{1}{4}\\ & \therefore y=\frac{1}{4}x-\frac{3}{4} \mbox{ and }y=-\frac{1}{4}x+\frac{3}{4}\\ \end{align}$
- $\begin{align} & \mbox{When } m=\frac{1}{4}\\ & \left[2\left(\frac{1}{16}\right)+1\right]x^{2}-12\left(\frac{1}{16}\right)x+18\left(\frac{1}{16}\right)-1=0\\ & \frac{9}{8}x^{2}-\frac{3}{4}x+\frac{1}{8}=0\\ & 9x^{2}-6x+1=0\\ & \left(3x-1\right)^{2}=0\\ & x=\frac{1}{3}, y=\frac{1}{4}\left(\frac{1}{3}\right)-\frac{3}{4}=-\frac{2}{3}\\ & \mbox{When } m=-\frac{1}{4}\\ & \left[2\left(\frac{1}{16}\right)+1\right]x^{2}-12\left(\frac{1}{16}\right)x+18\left(\frac{1}{16}\right)-1=0\\ & \therefore x=\frac{1}{3}, y=-\frac{1}{4}\left(\frac{1}{3}\right)+\frac{3}{4}=\frac{2}{3}\\ & \therefore \mbox{points of contact are } \left(\frac{1}{3},-\frac{2}{3}\right), \left(\frac{1}{3},\frac{2}{3}\right) \end{align}$

c) passes through $\left(4,6 \right)$ . $x^{2}+12y^{2}=48\,$ .

$\begin{align} & y=\frac{1}{2}x+4 \mbox{ and }y=-2x+14\\ & \left(-6,1\right), \left(\frac{48}{7},\frac{2}{7}\right)\\ \end{align}$
- $\begin{align} & \mbox{Let tangent be } y=mx+c \frac{\qquad}{}(1)\\ & x^{2}+12y^{2}=48 \frac{\qquad}{}(2)\\ & (1) \to (2) : x^{2}+12\left(mx+c\right)^{2}=48 \\ & x^{2}+12\left(m^{2}x^{2}+2mcx+c^{2}\right)-48=0\\ & \left(12m^{2}+1\right)x^{2}+24mcx+12\left(c^{2}-4\right)=0\\ & \mbox{Since tangent,} \therefore \mbox{equal roots}\\ & \therefore b^{2}-4ac=0\\ & \left(-24mc\right)^{2}-48\left(12m^{2}+1\right)\left(c^{2}-4\right)=0\\ & 576m^{2}c^{2}-48\left(12m^{2}c^{2}-48m^{2}+c^{2}-4\right)=0\\ & -48m^{2}+c^{2}-4=0\\ & 48m^{2}-c^{2}+4=0\frac{\qquad}{}(3)\\ & y=mx+c \\ & \left(4,6 \right) : 6=4m+c, \therefore c=6-4m\frac{\qquad}{}(4)\\ & (4)\to(3):\\ & 48m^{2}-\left(36-48m+16m^{2}\right)+4=0 \\ & 32m^{2}+48m-32=0 \\ & 2m^{2}+3m-2=0\\ & \left(2m-1\right)\left(m+2\right)=0\\ & m=\frac{1}{2} \mbox{ or } m=-2 \\ & \mbox{When }m=\frac{1}{2}, c= 4; m=-2, c=14\\ & \therefore y=\frac{1}{2}x+4 \mbox{ and }y=-2x+14\\ \end{align}$
- $\begin{align} & \mbox{When } m=\frac{1}{2}, c=4\\ & \left[12\left(\frac{1}{4}\right)+1\right]x^{2}+24\left(\frac{1}{2}\right)\left(4\right)x+12\left(4^{2}-4\right)=0\\ & 4x^{2}+48x+144=0\\ & x^{2}+12x+36=0\\ & \left(x+6\right)^{2}=0\\ & x=-6, y=\frac{1}{2}\left(-6\right)+4=1\\ & \mbox{When } m=-2, c=14\\ & \left[12\left(-2\right)^{2}+1\right]x^{2}+24\left(-2\right)\left(14\right)x+12\left(14^{2}-4\right)=0\\ & 49x^{2}-672x+2304=0\\ & \left(7x-48\right)^{2}=0\\ & x=\frac{48}{7}, y=-2\left(\frac{48}{7}\right)+14=\frac{2}{7}\\ & \therefore \mbox{points of contact are } \left(-6,1\right), \left(\frac{48}{7},\frac{2}{7}\right) \end{align}$

4. a) Prove that $x+3y-11=0\,$ is tangent to the ellipse $x^{2}+2y^{2}=22\,$ . Find the point of contact.

$\left(2,3\right)\,$
- $\begin{align} & x+3y-11=0 \\ & \therefore x=11-3y\frac{\qquad}{}(1)\\ & x^{2}+2y^{2}=22 \frac{\qquad}{}(2)\\ & (1) \to (2) : \left(11-3y\right)^{2}+2y^{2}=22 \\ & 121-66y+9y^{2}+2y^{2}=22 \\ & 11y^{2}-66y+99=0 \\ & y^{2}-6y+9=0 \\ & \left(y-3\right)^{2}=0 \\ & y=3 \mbox{ (repeated)}\\ & \mbox{Since repeated roots }, \therefore\mbox{tangent}\frac{\qquad}{}\mbox{(proved)}\\ & y=3, x=2 \\ & \therefore \mbox{Point of contact is }\left(2,3\right) \\ \end{align}$

b) Determine whether $y=3x+7\,$ is tangent to $x^{2}+4y^{2}=8\,$ .

$\mbox{No}\,$
- $\begin{align} & y=3x+7\frac{\qquad}{}(1)\\ & x^{2}+4y^{2}=8 \frac{\qquad}{}(2)\\ & (1) \to (2) : x^{2}+4\left(3x+7\right)^{2}=8 \\ & x^{2}+4\left(9x^{2}+42x+49\right)-8=0 \\ & 37x^{2}+168x+188=0\\ &\left(x+2\right)\left(37x-94\right)=0\\ & \mbox{Since not repeated roots }, \therefore\mbox{not tangent}\\ \end{align}$

5. If the line $y=mx+c\,$ touches the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , prove that $a^{2}m+b^{2}=c^{2}\,$ .

- $\begin{align} & y=mx+c\frac{\qquad}{}(1)\\ & \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\\ & b^{2}x^{2}+a^{2}y^{2}=a^{2}b^{2} \frac{\qquad}{}(2)\\ & (1) \to (2) : b^{2}x^{2}+a^{2}\left(mx+c\right)^{2}=a^{2}b^{2}\\ & b^{2}x^{2}+a^{2}\left(m^{2}x^{2}+2mcx+c^{2}\right)-a^{2}b^{2}=0\\ & b^{2}x^{2}+a^{2}m^{2}x^{2}+2a^{2}mcx+a^{2}c^{2}-a^{2}b^{2}=0\\ & \left(a^{2}m^{2}+b^{2}\right)x^{2}+2a^{2}mcx+a^{2}c^{2}-a^{2}b^{2}=0\\ & \mbox{Since tangent }, \therefore\mbox{repeated roots}\\ & b^{2}-4ac=0 \\ & \left(2a^{2}mc\right)^{2}-4\left(a^{2}m^{2}+b^{2}\right)\left[a^{2}\left(c^{2}-b^{2}\right)\right] =0 \\ & 4a^{4}m^{2}c^{2}-4a^{2}\left(a^{2}m^{2}c^{2}-a^{2}m^{2}b^{2}+b^{2}c^{2}-b^{4}\right)=0\\ & \therefore -a^{2}m^{2}b^{2}+b^{2}c^{2}-b^{4} =0 \\ & a^{2}m^{2}-c^{2}+b^{2}=0 \\ & \therefore a^{2}m^{2}+b^{2}=c^{2}\frac{\qquad}{}\mbox{(proved)} \\ \end{align}$

6. Determine whether the points $\left(2,2\right)$ , $\left(-2,1\right)$ , $\left(1,2\right)$ lies within or outside the ellipse $x^{2}+2y^{2}=9\,$

$\left(2,2\right)\mbox{outside},\left(-2,1\right)\mbox{within},\left(1,2\right)\mbox{on}$
- $\begin{align} & \mbox{Let tangent be } y=mx+c \frac{\qquad}{}(1)\\ & x^{2}+2y^{2}=9 \frac{\qquad}{}(2)\\ & (1) \to (2) : x^{2}+2\left(mx+c\right)^{2}=9 \\ & x^{2}+2\left(m^{2}x^{2}+2mcx+c^{2}\right)-9=0\\ & \left(2m^{2}+1\right)x^{2}+4mcx+2c^{2}-9=0\\ & \mbox{Since tangent,} \therefore \mbox{equal roots}\\ & \therefore b^{2}-4ac=0\\ & \left(4mc\right)^{2}-4\left(2m^{2}+1\right)\left(2c^{2}-9\right)=0\\ & 16m^{2}c^{2}-4\left(4m^{2}c^{2}-18m^{2}+2c^{2}-9\right)=0\\ & -18m^{2}+2c^{2}-9=0\\ & 18m^{2}-2c^{2}+9=0\frac{\qquad}{}(3)\\ \end{align}$
- $\begin{align} & y=mx+c \\ & \left(2,2 \right) : 2=2m+c, \therefore c=2-2m\frac{\qquad}{}(4)\\ & (4)\to(3):\\ & 18m^{2}-2\left(4-8m+4m^{2}\right)+9=0 \\ & 14m^{2}+16m+1=0 \\ & \mbox{For } 14m^{2}+16m+1=0 \\ & b^{2}-4ac=256-19=237>0 \\ & \therefore \mbox{Two distinct real values of } m \\ & \therefore \mbox{Two tangents can be drawn from point}\\ \end{align}$
- Media:MT-CG-E7-6i.dia
- $\therefore \left(2,2\right)\mbox{lies outside the ellipse}$
- $\begin{align} & y=mx+c \\ & \left(-2,1 \right) : 1=-2m+c, \therefore c=1+2m\frac{\qquad}{}(5)\\ & (5)\to(3):\\ & 18m^{2}-2\left(1+4m+4m^{2}\right)+9=0 \\ & 14m^{2}-8m+7=0 \\ & \mbox{For } 14m^{2}-8m+7=0 \\ & b^{2}-4ac=64-392=-328<0 \\ & \therefore \mbox{No real values of } m \\ & \therefore \mbox{No tangent can be drawn from point}\\ \end{align}$
- Media:MT-CG-E7-6ii.dia
- $\therefore \left(-2,1\right)\mbox{lies within the ellipse}$
- $\begin{align} & y=mx+c \\ & \left(1,2 \right) : 2=m+c, \therefore c=2-m\frac{\qquad}{}(6)\\ & (6)\to(3):\\ & 18m^{2}-2\left(4-4m+m^{2}\right)+9=0 \\ & 16m^{2}+8m+1=0 \\ & \left(4m+1\right)\\ & \therefore \mbox{One real value of } m \\ & \therefore \mbox{Only one tangent can be drawn from point}\\ \end{align}$
- Media:MT-CG-E7-6iii.dia
- $\therefore \left(-2,1\right)\mbox{lies on the ellipse}$

Exercise 8

1. Sketch the graph of the following parabolas

a) $y^{2}=6x\,$

- Media:MT-CG-E8-1a.dia
- $\mbox{Vertex }= \left(0,0\right)$

b) $y^{2}=8x-4\,$

- Media:MT-CG-E8-1b.dia
- $\begin{align} & y^{2}=8x-4 \\ & y^{2}=8\left(x-\frac{1}{2}\right) \\ & \mbox{Vertex }= \left(\frac{1}{2},0\right) \end{align}$

c) $y^{2}=\frac{1}{3}x-1\,$

- Media:MT-CG-E8-1c.dia
- $\begin{align} & y^{2}=\frac{1}{3}x-1 \\ & y^{2}=\frac{1}{3}\left(x-3\right) \\ & \mbox{Vertex }= \left(3,0\right) \end{align}$

d) $y^{2}=-4x\,$

- Media:MT-CG-E8-1d.dia
- $\mbox{Vertex }= \left(0,0\right)$

e) $y^{2}=-x+3\,$

- Media:MT-CG-E8-1e.dia
- $\begin{align} & y^{2}=-x+3 \\ & y^{2}=-\left(x-3\right) \\ & \mbox{Vertex }= \left(3,0\right)\\ & \mbox{When } x=0, y=\pm\sqrt{3} \end{align}$

f) $\left(y-1\right)^{2}=x+2\,$

- Media:MT-CG-E8-1f.dia
- $\begin{align} & \left(y-1\right)^{2}=x+2 \\ & \mbox{Vertex }= \left(-2,1\right)\\ & \mbox{When } x=0, y=1\pm\sqrt{2} & \mbox{When } y=0, x=-1 \end{align}$

g) $y^{2}+4y=2x-4\,$

- Media:MT-CG-E8-1g.dia
- $\begin{align} & y^{2}+4y=2x-4 \\ & \left(y+2\right)^{2}=2x \\ & \mbox{Vertex }= \left(0,-2\right)\\ & \mbox{When } y=0, x=4 \end{align}$

h) $x^{2}=2y\,$

- Media:MT-CG-E8-1h.dia
- $\mbox{Vertex }= \left(0,0\right)$

i) $\left(x-2\right)^{2}=y+4$

- Media:MT-CG-E8-1i.dia
- $\begin{align} & \left(x-2\right)^{2}=y+4 \\ & \mbox{Vertex }= \left(2,-4\right)\\ & \mbox{When } x=0, y=0\\ & \mbox{When } y=0, x=0 \mbox{ or } -4 \\ \end{align}$

2. Sketch the graph of the following hyperbolas

a) $x^{2}-y^{2}=1\,$

- Media:MT-CG-E8-2a.dia
- $\begin{align} & x^{2}-y^{2}=1 \\ & \mbox{When } y=0, x=\pm1\\ & y^{2}=x^{2}-1\\ & \mbox{As } x \to \infty, y^{2}\to x^{2},\therefore y \to \pm x \\ & \therefore \mbox{Asymptotes }: y=x, y=-x\\ \end{align}$

b) $x^{2}-16y^{2}=4\,$

- Media:MT-CG-E8-2b.dia
- $\begin{align} & x^{2}-16y^{2}=4 \\ & \mbox{When } y=0, x=\pm 2 \\ & 16y^{2}=x^{2}-4\\ & \mbox{As } x \to \infty, 16y^{2}\to x^{2},\therefore y \to \pm \frac{1}{4}x \\ & \therefore \mbox{Asymptotes }: y=\frac{1}{4}x, y=-\frac{1}{4}x\\ \end{align}$

c) $y^{2}-\frac{x^{2}}{25}=1\,$

- Media:MT-CG-E8-2c.dia
- $\begin{align} & y^{2}-\frac{x^{2}}{25}=1 \\ & \mbox{When } x=0, y=\pm 1 \\ & y^{2}=\frac{x^{2}}{25}+1\\ & \mbox{As } x \to \infty, y^{2}\to \frac{x^{2}}{25},\therefore y \to \pm \frac{1}{5}x \\ & \therefore \mbox{Asymptotes }: y=\frac{1}{5}x, y=-\frac{1}{5}x\\ \end{align}$

3. Sketch the graph of the following hyperbolas

a) $xy=5\,$

- Media:MT-CG-E8-3a.dia
- $\begin{align} & \mbox{Asymptotes }: y=0, x=0\\ \end{align}$

b) $3xy=-1\,$

- Media:MT-CG-E8-3b.dia
- $\begin{align} & \mbox{Asymptotes }: y=0, x=0\\ \end{align}$

c) $x\left(y+2\right)=8\,$

- Media:MT-CG-E8-3c.dia
- $\begin{align} & \mbox{Asymptotes }: y=-2, x=0\\ \\ & \mbox{When } y=0, x=4 \\ \end{align}$

d) $\left(x-2\right)\left(y+1\right)=4\,$

- Media:MT-CG-E8-3d.dia
- $\begin{align} & \mbox{Asymptotes }: y=-1, x=+2\\ \\ & \mbox{When } y=0, x=6 \\ & \mbox{When } x=0, y=-3 \\ \end{align}$

e) $x+1=\frac{-3}{2y-4}$

- Media:MT-CG-E8-3e.dia
- $\begin{align} & \mbox{Asymptotes }: y=2, x=-1\\ \\ & \mbox{When } y=0, x=-\frac{1}{4} \\ & \mbox{When } x=0, y=\frac{1}{2} \\ \end{align}$

Coordinate Geometry Part2

From StpmWiki

Contents

Exercise 4

Exercise 5

Exercise 6

Exercise 7

Exercise 8

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