Let A be a point on the x -axis. Common tangents are drawn from A...

Let a tangent on parabola ${y}^{2} = 16x$ be,

$y = mx + \dfrac{4}{m}$

It is also common to ${x}^{2} + {y}^{2} = 8$

$\text{If}{\left(\dfrac{4}{m}\right)}^{2} = 8\left(1 + {m}^{2}\right)$

$⇒ \dfrac{16}{{m}^{2}} = 8 ( 1 + {m}^{2} )$

$⇒ {m}^{4} + {m}^{2} - 2 = 0 ⇒ ( {m}^{2} + 2 ) ( {m}^{2} - 1 ) = 0$

$⇒ m = \pm 1$

Taking one of the tangents $y = x + 4$

$⇒ x - y + 4 = 0$

Tangent to ${x}^{2} + {y}^{2} = 8$ is $x{x}_{1} + y{y}_{1} - 8 = 0$

$∴ \dfrac{{x}_{1}}{1} = \dfrac{{y}_{1}}{1} = \dfrac{- 8}{4}$

$⇒ Q ( - 2, 2 )$

Tangent to ${y}^{2} = 16x$ is $y{y}_{1} = 8 ( x + {x}_{1} )$

$⇒ 8x - y{y}_{1} + 8{x}_{1} = 0$

$∴ \dfrac{8}{1} = \dfrac{- {y}_{1}}{- 1} = \dfrac{8{x}_{1}}{4}$

$⇒ {y}_{1} = 8, {x}_{1} = 4$

$∴ R ( 4, 8 )$

Using the distance formula between two points,

$QR = \sqrt{( 4 + 2{)}^{2} + ( 8 - 2{)}^{2}} = 6\sqrt{2}$

$∴ ( QR{)}^{2} = {\left(6\sqrt{2}\right)}^{2} = 72$