The diameters of a circle are along $2 x+y-7=0$ and $x+3 y-11=0$. Then, the equation of this circle, which…
The diameters of a circle are along $2 x+y-7=0$ and $x+3 y-11=0$. Then, the equation of this circle, which also passes through $(5,7)$, is
$x^2+y^2-4 x-6 y-16=0$
$x^2+y^2-4 x-6 y-20=0$
$x^2+y^2-4 x-6 y-12=0$
$x^2+y^2+4 x+6 y-12=0$
Solution
The intersection point of diameter lines is $(2,3)$ which is the centre of circle.
Now, radius
$\begin{aligned}
& =\sqrt{(5-2)^2+(7-3)^2} \\
& =\sqrt{9+16}=5
\end{aligned}$
$\therefore$ Required equation of circle is
$\begin{aligned}
(x-2)^2+(y-3)^2 & =5^2 \\
\Rightarrow \quad x^2+y^2-4 x-6 y-12 & =0
\end{aligned}$