The equation of the circle having the common chord of the circles \(x^2+y^2-8 x=0\) and \(x^2+y^2-9=0\) as…
The equation of the circle having the common chord of the circles \(x^2+y^2-8 x=0\) and \(x^2+y^2-9=0\) as its diameter is
\(x^2+y^2-72 x-207=0\)
\(x^2+y^2+72 x+207=0\)
\(32 x^2+32 y^2-72 x-207=0\)
\(32 x^2+32 y^2+72 x-207=0\)
Solution
The equation of circle passes through the intersection of circles \(x^2+y^2-8 x=0\) and
\(x^2+y^2-9=0 \text { is }\)
\(\begin{aligned}
& \left(x^2+y^2-8 x\right)+\lambda\left(x^2+y^2-9\right)=0 \\
\Rightarrow \quad & (1+\lambda) x^2+(1+\lambda) y^2-8 x-9 \lambda=0 \\
\Rightarrow & x^2+y^2-\frac{8}{1+\lambda} x-9 \frac{\lambda}{1+\lambda}=0 \text { having centre } \\
C & \left(\frac{4}{1+\lambda}, 0\right)
\end{aligned}\)
Now, equation of common chord of given circles is \(8 x=9\), and the common chord is the diameter of the circle having centre \(C\left(\frac{4}{1+\lambda}, 0\right)\), so
\(\begin{aligned}
32 & =9+9 \lambda \\
\Rightarrow \quad \lambda=\frac{23}{9} \text { and }(1+\lambda) & =\frac{32}{9}
\end{aligned}\)
\(\therefore\) Equation of required circle is
\(\begin{aligned}
\frac{32}{9} x^2+\frac{32}{9} y^2-8 x-23 & =0 \\
\Rightarrow 32 x^2+32 y^2-72 x-207 & =0
\end{aligned}\)
Hence, option (c) is correct.