Q93 / 160 · TS EAMCET 2019 (03 May Shift 1) · Mathematics
The equation of the circle whose diameter is the common chord of the circles $x^2+y^2-3 x+y-10=0$ and $x^2+y^2-x+2 y-20=0$, is
- A$x^2+y^2-3 x+6 y+15=0$
- B$x^2+y^2-6 x+4 y+10=0$
- C$x^2+y^2-9 x+2 y+20=0$
- Correct answerD$x^2+y^2-9 x-2 y+20=0$
Solution
Equation of given circles
$$
\begin{aligned}
& S_1=x^2+y^2-3 x+y-10=0 \\
& S_2=x^2+y^2-x+2 y-20=0
\end{aligned}
$$
$\because$ Equation of family of circles passes through intersection of circles $S_1$ and $S_2$ is
$$
\begin{aligned}
& S_1+\lambda S_2=0 \\
& x^2(1+\lambda)+y^2(1+\lambda)-x(3+\lambda)+y(1+2 \lambda) \\
&-10-20 \lambda=0
\end{aligned}
$$
$\because$ Equation of common chord of circles $S_1$ and $S_2$
$$
S_1-S_2=-2 x-y+10=0
$$
$2 x+y-10=0$ is diametre of circle $S_1+\lambda S_2=0$
Centre of circle $S_1+\lambda s_2=0$ is lie on
$$
\begin{aligned}
& 2 x+y-10=0 \\
& \because \text { Centre }\left(\frac{3+\lambda}{2(1+\lambda)},-\frac{(1+2 \lambda)}{2(1+\lambda)}\right)
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad & 2\left(\frac{3+\lambda}{2(1+\lambda)}\right)-\left(\frac{1+2 \lambda}{2(1+\lambda)}\right)-10=0 \\
& 6+2 \lambda-1-2 \lambda-20-20 \lambda=0 \\
& \lambda=\frac{-3}{4}
\end{aligned}
$$
Putting the value of $\lambda$ in Eq. (i), we get
$$
\begin{aligned}
& x^2\left(1-\frac{3}{4}\right)+y^2\left(1-\frac{3}{4}\right)-x\left(3-\frac{3}{4}\right)+y\left(1-\frac{6}{4}\right) \\
& -10+\left(20 \times \frac{3}{4}\right)=0 \\
& \Rightarrow x^2+y^2-9 x-2 y+20=0
\end{aligned}
$$
Hence, the correct answer is option: 'D'.