Q92 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics

A circle $S=x^2+y^2+2 g x+2 f y+4=0$ cuts the circle $x^2$ $+y^2-4 x-4 y-4=0$ orthogonally and makes an angle of $60^{\circ}$ with the circle $x^2+y^2+4 x+4 y+4=0$. Then the radius of the circle $\mathrm{S}=0$ is

  • Correct answer
    A
    4
  • B
    3
  • C
    5
  • D
    1

Solution

2 circles cut orthogonally then $$ \begin{aligned} & 2 g_1 g_2+2 f_1 f_2=C_1+C_2 \\ \Rightarrow & 2 g(-2)+2 f(-2)=4-4 \\ \Rightarrow & g+f=0 \end{aligned} $$ Also angle between circles, $$ \begin{aligned} & \cos \theta=\frac{r_1^2+r_2^2-d^2}{2 r_2} \\ \Rightarrow & \cos 60^{\circ}=\frac{\eta^2+4-(2-g)^2-(2-f)^2}{2 r_1 \cdot(2)} \\ \Rightarrow & 2 \eta_1=\left|\eta_1^2+4-\left(4+g^2-4 g\right)-\left(4+f^2-4 f\right)\right| \\ \Rightarrow & \left.2 r_1=\mid \eta^2+4-4-g^2+4 g-4-f^2+4 f\right) \mid \\ \Rightarrow & 2 r_1=\left|\eta_1^2-g^2-f^2-4+4(g+f)\right| \\ \Rightarrow & 2 r_1=\left|\eta_1^2-\left(g^2+f^2\right)-4\right| \\ \Rightarrow & 2 r_1=\left|\eta_1^2-\left(g^2+f^2-4\right)-8\right| \\ \Rightarrow & 2 r_1=\left|r_1^2-r_1^2-8\right| \Rightarrow 2 \eta_1=8 \Rightarrow \eta=4 . \end{aligned} $$

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

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