Let C be a circle passing through the points A(2,-1) and B(3,4).

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 $(x-5)^2+(y-1)^2=\dfrac{13}2$, then $r^2$ is equal to:

Accepted Answer

Equation of perpendicular bisector of AB is

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

Solving it with equation of given circle,

$${(x - 5)^2}{\left( {{{10 - x} \over 5} - 1} \right)^2} = {{13} \over 2}$$

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

$$ \Rightarrow x - 5 =  \pm \,{5 \over 2} \Rightarrow x = {5 \over 2}$$ or $${{15} \over 2}$$

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

So, centre will be $$\left( {{{15} \over 2},{1 \over 2}} \right)$$

Now, $${r^2} = {\left( {{{15} \over 2} - 2} \right)^2} + {\left( {{1 \over 2} + 1} \right)^2}$$

$$ = {{65} \over 2}$$

Solutions · 1