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Let $C$ be a circle passing through the points $A(2,-1)$ and $B(3,4)$ . The line segment $A B$ is not a diameter of $\mathrm{C}.$ If $\mathrm{r}$ is the radius of $\mathrm{C}$ and its centre lies on the circle $(x-5)^{2}+(y-1)^{2}=\frac{13}{2}$ , then $r^{2}$ is equal to:
Option: 1
$32$

Option: 2
$\frac{65}{2}$

Option: 3
$\frac{61}{2}$

Option: 4
$30$

Answers (1)

best_answer

$\mathrm{A B =\sqrt{26}} \\$

$\mathrm{r^{2} =CM^{2}+AM^{2}} \\$

$\mathrm{=\left(2.\sqrt{\frac{13}{2}}\right)^{2}+\left(\sqrt{\frac{13}{2}}\right)^{2}} \\$

$\mathrm{r^{2} =\frac{65}{2}}$

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