Find the equation of a circle which passes through the point $(1,2)$ and the points of intersection of the…
Find the equation of a circle which passes through the point $(1,2)$ and the points of intersection of the circles $x^2+y^2-8 x-6 y+21=0$ and $x^2+y^2-2 x-15=0$
$x^2+y^2-6 x-4 y+9=0$
$x^2+y^2-18 x-12 y+27=0$
$2\left(x^2+y^2\right)-18 x+12 y+27=0$
$4\left(x^2+y^2\right)-3 x+12 y+16=0$
Solution
$
\begin{gathered}
C_1: x^2+y^2-8 x-6 y+21=0 \\
C_2: x^2+y^2-2 x-15=0
\end{gathered}
$
Equation of circle passing through the point of intersection.
$
\begin{gathered}
S_1+\lambda S_2=0 \\
\left(x^2+y^2-8 x-6 y+21\right)+s\left(x^2+y^2-2 x-15\right)=0
\end{gathered}
$
Now, equation of circle passing through $(1,2)$ is
$
\begin{aligned}
& \left(1^2+2^2-8(1)-6(2)+21\right) \\
& \left.+\lambda\left\{1(1)^2+(2)^2-2(1)-15\right)\right\}=0 \\
& \Rightarrow(1+4-8-12+21)+\lambda(5-2-15)=0 \\
& \Rightarrow 6+\lambda(-12)=0 \Rightarrow \lambda=\frac{1}{2}
\end{aligned}
$
$\therefore$ Equation of required circle is
$
\begin{aligned}
& \left(x^2+y^2-8 x-6 y+21\right)+\frac{1}{2}\left(x^2+y^2-2 x-15\right)=0 \\
& \Rightarrow \quad 3 x^2+3 y^2-18 x-12 y+42-15=0 \\
& \Rightarrow \quad 3\left(x^2+y^2-6 x-4 y+9\right)=0 \\
& \therefore \quad x^2+y^2-6 x-4 y+9=0 \\
&
\end{aligned}
$