JEE Main 2022 — Mathematics Coordinate Geometry

Equation of line joining $AB$AB is

$$y+1=5(x-2)$$y+1=5(x−2)

i.e. $5x-y-11=0$5x−y−11=0

Also $AB=\sqrt{{(3-2)}^{2}+{(4+1)}^{2}}=\sqrt{26}$AB=(3−2)2+(4+1)2​=26​

$r$r is the radius of the circle $C$C

For the circle ${(x-5)}^{2}+{(y-1)}^{2}=\frac{13}{2}$(x−5)2+(y−1)2=213​

centre is $(5,1)$(5,1) and radius $R=\sqrt{\frac{13}{2}}$R=213​​

Equation of perpendicular bisector of $AB$AB will be $y-1=-\frac{1}{5}(x-5)\Rightarrow x+5y-10=0$y−1=−51​(x−5)⇒x+5y−10=0

Solving this equation of line with ${(x-5)}^{2}+{(y-1)}^{2}=\frac{13}{2}$(x−5)2+(y−1)2=213​, we get

$M$M as $(\frac{5}{2},\frac{3}{2})$(25​,23​)

So $M$M lies on the line joining $AB$AB.

Now ${r}^{2}=C{M}^{2}+A{M}^{2}$r2=CM2+AM2

$$={(2\times \sqrt{\frac{13}{2}})}^{2}+{(\sqrt{\frac{13}{2}})}^{2}$$=(2×213​​)2+(213​​)2

$${r}^{2}=\frac{65}{2}$$r2=265​