Q92 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Mathematics

$\left(0, \frac{3}{4}\right)$ is the radical centre of the circles $\begin{aligned} & S_1: x^2+y^2-2 x+6 y=0, \\ & S_2: x^2+y^2+2 g x-2 y+6=0 \text { and } \\ & S_3: x^2+y^2-12 x+2 f y+3=0 . \text { If } S_2 \text { and } S_3 \end{aligned}$ intersect orthogonally, then $(g, f)=$

  • Correct answer
    A
    $\left(\frac{-11}{12}, 1\right)$
  • B
    $\left(1, \frac{-21}{2}\right)$
  • C
    $\left(0, \frac{-9}{2}\right)$
  • D
    $\left(-1, \frac{-7}{12}\right)$

Solution

Radical axis of circle of $\begin{aligned} & S_1=x^2+y^2-2 x+6 y=0 \\ & S_2=x^2+y^2+2 g x-2 y+6=0, \\ & S_3=x^2+y^2-12 x+2 f y+3=0 \text { is } S_2-S_3=0 \\ & \therefore(2 g+12) x-(2+2 f) y+6-3=0 \\ & (2 g+12) x-(2+2 f) y+3=0 \end{aligned}$ $\left(0, \frac{3}{4}\right)$ is centre of radical axis $\begin{aligned} \therefore \quad & \left(0, \frac{3}{4}\right) \text { lie on } S_2-S_3=0 \\ & 0-(2+2 f) \frac{3}{4}+3=0 \Rightarrow 2+2 f=4 \Rightarrow f=1 \end{aligned}$ $S_2$ and $S_3$ intersect orthogonally $\begin{aligned} & \therefore \quad 2 g(-6)+2(-f)=6+3 \\ & -12 g-2 f=9 \quad \Rightarrow-12 g-2=9 \quad[\because f=1] \\ & \therefore \quad g=\frac{11}{-12}=\frac{-11}{12} \quad \therefore \quad(g, f)=\left(\frac{-11}{12}, 1\right) \\ & \end{aligned}$

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

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