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 r2 is equal to

• A. 32
• B. $\frac{65}{2}$
• C. $\frac{61}{2}$
• D. 30

Answer 1

Answer: (B)

$\frac{65}{2}$

Answer 2

The correct answer is (B) : $\frac{65}{2}$
Equation of perpendicular bisector of AB is
$y−32=−\frac{1}{5}(x−52)⇒x+5y=10$
Solving it with equation of given circle,
$(x−5)^2+(\frac{10−x}{5}−1)^2=\frac{13}{2}$
$⇒(x−5)^2(1+\frac{1}{25})=\frac{13}{2}$
$⇒x−5=±\frac{5}{2}⇒x=\frac{5}{2} or \frac{15}{2}$
But $x≠\frac{5}{2}$
because AB is not the diameter.
So, centre will be $(\frac{15}{2},\frac{1}{2})$
Now $r^2=(\frac{15}{2}−2)^2+(\frac{1}{2}+1)^2$
$=\frac{65}{2}$