Question

Chord of contact $\left(L\right)$ from point $P$ on the circle $x^2+y^2-2x-4y-20=0$ makes an angle of ${60}^{\circ }$ at a point on the circumference of circle. If slope of the chord of contact is $\frac{1}{2},$ then

• A. $L:x-2y=\frac{-(5\sqrt{5}+6)}{2}$
• B. $L:x-2y=\frac{5\sqrt{5}-6}{2}$
• C. $P\equiv (1+2\sqrt{5},2-4\sqrt{5})$
• D. $P\equiv (1-2\sqrt{5},2+4\sqrt{5})$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $L:x-2y=\frac{-(5\sqrt{5}+6)}{2}$
B $L:x-2y=\frac{5\sqrt{5}-6}{2}$
C $P\equiv (1+2\sqrt{5},2-4\sqrt{5})$
D $P\equiv (1-2\sqrt{5},2+4\sqrt{5})$


Let coordinates of $P$ be $(h,k).$ Then
$L:hx+ky-(x+h)-2(y+k)-20=0$
$\Rightarrow L:x(h-1)+y(k-2)-h-2k-20=0$
As $-\frac{h-1}{k-2}=\frac{1}{2},$
$\Rightarrow 2h+k=4\cdots \left(1\right)$

$OA=r=5$
$\cos {60}^{\circ }=\frac{r}{OP}$
$\therefore OP=10$
$\therefore 100=(h-1{)}^2+(k-2{)}^2$
$\Rightarrow 100=(h-1{)}^2+4(h-1{)}^2\left[\text{Using (1)}\right]$
$\Rightarrow (h-1{)}^2=20$
$\Rightarrow h=1\pm 2\sqrt{5}$
$\therefore (h,k)=(1\pm 2\sqrt{5},2\mp 4\sqrt{5})$

We have
$L:x(h-1)+y(k-2)-h-2k-20=0$
$\Rightarrow x\left(\frac{h-1}{k-2}\right)+y=\frac{h+2k+20}{k-2}$
$\Rightarrow x-2y=-2\left(\frac{h+2k+20}{k-2}\right)$
Putting the value of $(h,k)$ in equation of $L,$
$x-2y=-\frac{(6\pm 5\sqrt{5})}{2}$