Mathematics - Circle Question with Solution | TestHub

Let the centre be $\left(-2a,\frac{6a-2}{2}\right)\equiv (-2a,3a-1)$ Centre is equal distance from $(4,2)$ and $(0,2)$ $\begin{array}{rl}& \Rightarrow \sqrt{(4+2a{)}^2+(3a-3{)}^2}=\sqrt{(-2a-0{)}^2+(3a-3{)}^2}\\ & \Rightarrow (2a+4{)}^2+9(a-1{)}^2=4a^2+9(a-1{)}^2\\ & \Rightarrow 4a^2+16+16a=4a^2\Rightarrow a=-1\\ & \Rightarrow \text{ centre }\equiv (2,-4)\Rightarrow \text{ Radius }=\sqrt{40}\end{array}$
$\begin{array}{rl}& \Rightarrow A{M}^2=(\sqrt{40}{)}^2-(\sqrt{37}{)}^2\\ & \Rightarrow 2AM=AB=2\sqrt{3}\end{array}$