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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 incircle of the $\triangle P Q R$ is
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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 incircle of the $\triangle P Q R$ is
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Verified Answer
The correct answer is:
2
2
Radius of incircle is, $r=\frac{\Delta}{s}$
As $\Delta=16 \sqrt{2}$
$$
\begin{aligned}
& \therefore \quad s=\frac{6 \sqrt{2}+6 \sqrt{2}+4 \sqrt{2}}{2}=8 \sqrt{2} \\
& \therefore \quad r=\frac{16 \sqrt{2}}{8 \sqrt{2}}=2 \\
&
\end{aligned}
$$
As $\Delta=16 \sqrt{2}$
$$
\begin{aligned}
& \therefore \quad s=\frac{6 \sqrt{2}+6 \sqrt{2}+4 \sqrt{2}}{2}=8 \sqrt{2} \\
& \therefore \quad r=\frac{16 \sqrt{2}}{8 \sqrt{2}}=2 \\
&
\end{aligned}
$$
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