Question

Let $C$ be a circle passing through the points $A(2,-1)$ and $B(3,4)$. The line segment $AB$ is not a diameter of $C$. If $r$ is the radius of $C$ and its centre lies on the circle $(x-5{)}^2+(y-1{)}^2=\frac{13}{2}$, then $r^2$ is equal to :

Answer 1

Equation of perpendicular bisector of $AB$ is

$y-\frac{3}{2}=-\frac{1}{5}(x-\frac{5}{2})\Rightarrow x+5y=10$

Solving it with equation of given circle,

$(x-5{)}^2+{(\frac{10-x}{5}-1)}^2=\frac{13}{2}$

$\Rightarrow (x-5{)}^2(1+\frac{1}{25})=\frac{13}{2}$

$\Rightarrow x-5=\pm \frac{5}{2}\Rightarrow x=\frac{5}{2}\text{or}\frac{15}{2}$

But $x\ne \frac{5}{2}$ because $AB$ is not the diameter.

So, centre will be $(\frac{15}{2},\frac{1}{2})$

Now, $r^2={(\frac{15}{2}-2)}^2+{(\frac{1}{2}+1)}^2=\frac{65}{2}$