Q93 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Mathematics

The equation of the circle passing through the points of intersection of the two orthogonal circles $\begin{aligned} & S_1=x^2+y^2+k x-4 y-1=0, \\ & S_2=3 x^2+3 y^2-14 x+23 y-15=0 \text { and passing } \end{aligned}$ through the point $(-1,-1)$ is

  • A
    $x^2+y^2-8 x-2 y-12=0$
  • B
    $3 x^2+3 y^2+18 x-12 y=0$
  • Correct answer
    C
    $5 x^2+5 y^2-22 x+15 y-17=0$
  • D
    $x^2+y^2-5 x+14 y+7=0$

Solution

We have, $\begin{aligned} S_1: x^2+y^2+k x-4 y-1 & =0 \\ S_2: 3 x^2+3 y^2-14 x+23 y-15 & =0 \\ =x^2+y^2-\frac{14}{3} x+\frac{23}{3} y-5 & =0 \end{aligned}$ Since, $S_1$ and $S_2$ are orthogonal $\therefore \quad 2\left[\left(\frac{-k}{2}\right)\left(\frac{7}{3}\right)+(2)\left(\frac{-23}{6}\right)\right]=-1-5$ $\begin{array}{ll}\Rightarrow & 2\left[\frac{-7 k}{6}-\frac{46}{6}\right]=-6 \Rightarrow \frac{-7 k}{6}-\frac{46}{6}=-3 \\ \Rightarrow & \frac{7 k}{6}=3-\frac{46}{6} \Rightarrow \frac{7 k}{6}=\frac{18-46}{6} \\ \Rightarrow & 7 k=-28 \Rightarrow k=-4 \\ \therefore & S_1: x^2+y^2-4 x-4 y-1=0 \\ & S_2: x^2+y^2-\frac{14}{3} x+\frac{23}{3} y-5=0\end{array}$ Equation of circle passing through intersection of $S_1$ and $S_2$ is $x^2+y^2-4 x-4 y-1+\lambda\left(x^2+y^2-\frac{14 x}{3}+\frac{23}{3} y-5\right)=0$ Since above circle passes through $(-1,-1)$, so $\begin{gathered} 1+1+4+4-1+\lambda\left(1+1+\frac{14}{3}-\frac{23}{3}-5\right)=0 \\ 9-6 \lambda=0 \Rightarrow \lambda=\frac{3}{2} \end{gathered}$ $\therefore$ Equation of required circle is $\begin{aligned} & x^2+y^2-4 x-4 y-1+\frac{3}{2}\left(x^2+y^2-\frac{14 x}{3}+\frac{23 y}{3}-5\right) \\ & =0 \\ & \Rightarrow \quad 2 x^2+2 y^2-8 x-8 y-2+3 x^2+3 y^2 \\ & \Rightarrow \quad 5 x^2+5 y^2-22 x+15 y-17=0 \end{aligned}$

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

Relevant Questions