Let $P$ be the point on the parabola $y = x^2$ such that the slope of the tangent to the parabola at the point $P$ is $4$ . Let Q be the point in the first quadrant lying on the circle ${x}^{2}+{y}^{2}=2$ such that the slope of the tangent to the circle at the point $Q$ is $-1$. Let $R$ be the point in the first quadrant lying on the ellipse ${x}^{2}+4{y}^{2}=8$ such that the slope of the tangent to the ellipse at the point $R$ is $−\frac{1}{2}$. Then the radius of the circle passing through the points $P, Q$ and $R$ is [JEE Advanced 2026]

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YESPYQ · Accepted Answer

Correct answer: c) $\sqrt{\frac{5}{2}}$. For $y=x^2$, slope $=2x$. $2x=4\Rightarrow x=2$, so $P=(2,4)$. For $x^2+y^2=2$, slope $=-\dfrac{x}{y}$. $-\dfrac{x}{y}=-1\Rightarrow x=y$. Since $x^2+y^2=2$, we get $2x^2=2\Rightarrow x=1$, so $Q=(1,1)$. For $x^2+4y^2=8$, slope $=-\dfrac{x}{4y}$. $-\dfrac{x}{4y}=-\dfrac12\Rightarrow x=2y$. $(2y)^2+4y^2=8\Rightarrow 8y^2=8\Rightarrow y=1$, so $R=(2,1)$. Let centre of circle be $C(h,k)$. Since $P=(2,4)$ and $R=(2,1)$, perpendicular bisector of $PR$ is $y=\dfrac52$. So centre is $C(h,\dfrac52)$. Using $CQ=CR$: $(h-1)^2+\left(\dfrac52-1\right)^2=(h-2)^2+\left(\dfrac52-1\right)^2$ $(h-1)^2=(h-2)^2\Rightarrow h=\dfrac32$ Radius $=CR=\sqrt{\left(\dfrac32-2\right)^2+\left(\dfrac52-1\right)^2}$ $r=\sqrt{\dfrac14+\dfrac94}=\sqrt{\dfrac{10}{4}}=\sqrt{\dfrac52}$

Let \(P\) be the point on the parabola \(y = x^2\) such that the slope of the tangent to the parabola at the point \(P\)…

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