Mathematics - Circle Question with Solution | TestHub

The centre of the circle is on the perpendicular bisector of the line joining $(-1,2)$ and $(1,2)$, which is the $y$-axis. The ordinate of the centre is given by $\begin{array}{l}(5-y{)}^2=1+(y-2{)}^2\\ \Rightarrow y=\frac{10}{3}\end{array}$ Hence, eq. of the circle is: $\begin{array}{l}{x}^2+{\left(y-\frac{10}{3}\right)}^2={\left(\frac{5}{3}\right)}^2\\ \Rightarrow 9x^2+9y^2-60y+75=0\end{array}$