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

The Correct Option is B

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 \(r^2=(\frac{15}{2}−2)^2+(\frac{1}{2}+1)^2\)
\(=\frac{65}{2}\)