The locus of the poles of the tangents to the circle $x^2+y^2-2 x+2 y-2=0$ with respect to the circle $x^2+y^2=4$, is

• Correct answer
  A

  $3 x^2+2 x y+3 y^2+8 x-8 y-16=0$
• B

  $x^2-2 x y+y^2-4 x+4 y+8=0$
• C

  $3 x^2-2 x y-3 y^2+4 x+4 y+16=0$
• D

  $x^2+y^2-4 x+4 y-8=0$

Solution

$C_1: x^2+y^2-2 x+2 y-2=0$ $C \equiv(1,-1), r=\sqrt{1^2+(-1)^2+2}=2$ $\Rightarrow(x-1)^2+(y+1)^2=4$ ...(i) $C_2: x^2+y^2=4$ ...(ii) $C \equiv(0,0), r=2$
If $P\left(x_1, y_1\right)$ i.e. pole lies outside of the circle then Chord of circle $=$ Locus of polar (i.e. $Q$ ) $\Rightarrow x x_1+y y_1-4=0$ ...(iii) This is the equation of polar. Perpendicular distance of point $O$ from straight line $Q S=2=\frac{\left|1 \times x_1-1 \times y_1-4\right|}{\sqrt{x_1^2+y_1^2}}$ Squaring both sides, we get : $\begin{aligned} & 4\left(x_1^2+y_1^2\right)=\left(x_1-y_1-4\right)^2 \\ & \Rightarrow \quad 4 x_1^2+4 y_1^2=x_1^2+y_1^2+16-2 x_1 y_1+8 y_1-8 x_1 \\ & \Rightarrow \quad 3 x_1^2+3 y_1^2+2 x_1 y_1+8 x_1-8 y_1-16=0\end{aligned}$ Taking locus of the points, we get : $3 x^2+3 y^2+2 x y+8 x-8 y-16=0$

Hence, the correct answer is option: 'A'.