JEE Main 2022 · 26 Jun · Shift 1 · Mathematics Q12

← JEE Main 2022 · 26 Jun · Shift 1

MathematicsSingle correct+4 / -1Question 12 of 90

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 :

\dfrac{65}2

\dfrac{61}2

Answer and worked solution

Official answer: B

Centre on the perpendicular bisector of AB: through $(2.5,1.5)$ with normal $(1,5)$ , i.e. $x+5y=10$ .

Intersect with $(x-5)^2+(y-1)^2=\frac{13}2$ : putting $x=10-5y$ gives $26(y-1)^2=\frac{13}2$ , so $y=\frac32$ or $\frac12$ . $y=\frac32$ is the midpoint of AB (excluded), so the centre is $\left(\frac{15}2,\frac12\right)$ .

Radius. $r^2=\left(\frac{11}2\right)^2+\left(\frac32\right)^2=\dfrac{65}2$ .

(B)

Topics:Equation of a circle Equations of a line

← Previous question Next question →

JEE Main 2022 · 26 Jun · Shift 1, Mathematics question 12