JEE Advanced 2020 · Paper 2 · Mathematics Q3

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MathematicsInteger+3 / -1Question 39 of 54

Let $O$ be the centre of the circle $x^2 + y^2 = r^2$ , where $r > \dfrac{\sqrt5}{2}$ . Suppose $PQ$ is a chord of this circle and the equation of the line passing through $P$ and $Q$ is $2x + 4y = 5$ . If the centre of the circumcircle of the triangle $OPQ$ lies on the line $x + 2y = 4$ , then the value of $r$ is ______

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Answer and worked solution

Official answer: 2

Locating the circumcentre. It lies on the perpendicular bisector of $PQ$ . Since $OP=OQ$ , that bisector passes through $O$ and is perpendicular to $2x+4y=5$ , i.e. it is the line $y=2x$ .

Meeting $x+2y=4$ gives the circumcentre $\left(\frac45,\frac85\right)$ .

The circumcircle. It passes through $O$ :
$x^2+y^2-\tfrac85x-\tfrac{16}5y=0.$

Common chord. Subtracting from $x^2+y^2=r^2$ gives the line through $P$ and $Q$ :
$\tfrac85x+\tfrac{16}5y=r^2\iff2x+4y=\tfrac54r^2.$
This must be $2x+4y=5$ , so $r^2=4$ and $r=2$ (which is indeed greater than $\frac{\sqrt5}2$ ).

Answer: 2

Topics:Equation of a circle Family of circles

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JEE Advanced 2020 · Paper 2, Mathematics question 3