JEE Advanced · Mathematics · 13. Parabola
Paragraph:
Consider the circle \(x^2+y^2=9\) and the parabola \(y^2=8 x\). They intersect at \(P\) and \(Q\) in the first and the fourth quadrants, respectively. Tangents to the circle at \(P\) and \(Q\) intersect the \(\mathrm{X}\)-axis at \(R\) and tangents to the parabola at \(P\) and \(Q\) intersect the \(\mathrm{X}\)-axis at \(S\).Question:
The radius of the circumcircle of the \(\triangle P R S\) is
- A
5
- B
\(3 \sqrt{3}\)
- C
\(3 \sqrt{2}\)
- D
\(2 \sqrt{3}\)
Answer & Solution
Correct Answer
(B)
\(3 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
Equation of circumcircle of \(\triangle P R S\) is
\[
(x+1)(x-9)+y^2+\lambda y=0
\]
It will pass through \((1,2 \sqrt{2})\), then
\[
\begin{aligned}
-16+8+\lambda \cdot 2 \sqrt{2} & =0 \\
\lambda \quad \lambda & =\frac{8}{2 \sqrt{2}}=2 \sqrt{2}
\end{aligned}
\]
Equation of circumcircle is
\[
x^2+y^2-8 x+2 \sqrt{2} y-9=0
\]
Hence, its radius is \(3 \sqrt{3}\).
ALITER
Let
\[
\begin{array}{ll}
\text { Let } & \angle P S R=\theta \\
\Rightarrow & \sin \theta=\frac{2 \sqrt{2}}{2 \sqrt{3}} \\
\Rightarrow & P R=6 \sqrt{2}=2 R \cdot \sin \theta
\end{array}
\]
\[
\Rightarrow \quad \sin \theta=\frac{2 \sqrt{2}}{2 \sqrt{3}}
\]
\[
\Rightarrow \quad R=3 \sqrt{3} \text {. }
\]
\[
(x+1)(x-9)+y^2+\lambda y=0
\]
It will pass through \((1,2 \sqrt{2})\), then
\[
\begin{aligned}
-16+8+\lambda \cdot 2 \sqrt{2} & =0 \\
\lambda \quad \lambda & =\frac{8}{2 \sqrt{2}}=2 \sqrt{2}
\end{aligned}
\]
Equation of circumcircle is
\[
x^2+y^2-8 x+2 \sqrt{2} y-9=0
\]
Hence, its radius is \(3 \sqrt{3}\).
ALITER
Let
\[
\begin{array}{ll}
\text { Let } & \angle P S R=\theta \\
\Rightarrow & \sin \theta=\frac{2 \sqrt{2}}{2 \sqrt{3}} \\
\Rightarrow & P R=6 \sqrt{2}=2 R \cdot \sin \theta
\end{array}
\]
\[
\Rightarrow \quad \sin \theta=\frac{2 \sqrt{2}}{2 \sqrt{3}}
\]
\[
\Rightarrow \quad R=3 \sqrt{3} \text {. }
\]
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