A triangle is formed by the tangents at the point ( 2, 2 ) on the...

Let

${S}_{1}: {y}^{2} = 2x$

${S}_{2}: {x}^{2} + {y}^{2} = 4x$

$P ( 2, 2 )$ is common point on ${S}_{1} \text{ and } {S}_{2}$

Let ${T}_{1}$ is tangent to ${S}_{1}$ at $P\left(2, 2\right)$, then

${T}_{1}: 2y = x + 2$

$⇒ {T}_{1}: x - 2y + 2 = 0$

And, ${T}_{2}$ is tangent to ${S}_{2}$ at $P\left(2, 2\right)$, then

${T}_{2}: 2x + 2y = 2\left(x + 2\right)$

$⇒ {T}_{2}: y = 2$

And, ${L}_{3}: x + y + 2 = 0$

$PQ = a = \sqrt{20}$

$QR = b = \sqrt{8}$

$RP = c = 6$

Also, ar\left(∆ PQR\right) = \dfrac{1}{2}\left|\left|\begin{cases}2 & 2 & 1 \\ - 2 & 0 & 1 \\ - 4 & 2 & 1\end{cases}\right|\right|

⇒ ar\left(∆ PQR\right) = \dfrac{1}{2}\left|\left|\begin{cases}2 & 2 & 1 \\ - 4 & - 2 & 0 \\ - 6 & 0 & 0\end{cases}\right|\right|

$⇒ ar\left(∆ PQR\right) = \dfrac{1}{2}\left|- 6\left(0 + 2\right)\right| = 6$

$∴ r = \dfrac{abc}{4\Delta} = \dfrac{\sqrt{160}}{4} = \sqrt{10}$

$⇒ {r}^{2} = 10$