Q88 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics

The power of a point $(2,0)$ with respect to a circle $S$ is -4 and the length of the tangent drawn from the point $(1,1)$ to $S$ is 2 . If the circle $S$ passes through the point $(-1,-1)$, then the radius of the circle $S$ is

  • A
    2
  • Correct answer
    B
    $\sqrt{13}$
  • C
    3
  • D
    $\sqrt{10}$

Solution

Let equation of circle be $x^2+y^2+2 g x+2 f y+C=0$ Since $(-1,-1)$ lies on circle $\therefore 1+1-2 g-2 f+C=0 \Rightarrow 2 g+2 f-C=2 \quad ...\mathrm{(i)}$ Since, power of a point $(2,0)$ with respect of circle $\begin{aligned} &=-4 \\ & \therefore(2)^2+0+4 g+0+C=0 \\ & \Rightarrow 4 g+C=-8 \Rightarrow C=-8-4 g \quad ...\mathrm{(ii)} \end{aligned}$ Length of tangent from $(1,1)$ is 2 . $\sqrt{1+1+2 g+2 f+C}=2 \Rightarrow 2 g+2 f+C=2$ From equation (ii) $2 g+2 f-8-4 g=2 \Rightarrow g-f=-5 \qquad ...\mathrm{(iii)}$ From equations (i) and (ii), we get $2 g+2 f+8+4 g=2 \Rightarrow 3 g+f=-3$ On solving equation (iii) and (iv), we get $\begin{aligned} & g=-2, f=3, C=0 \quad[\text { from (ii) }] \\ & \therefore \text { Radius }=\sqrt{g^2+f^2-C}=\sqrt{4+9+0}=\sqrt{13} \end{aligned}$

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

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