Let a common tangent to the curves y ^ 2 = 4x and (x - 4) ^ 2 + y ^...

Given,

Common tangent to the curves ${y}^{2} = 4x$ and ${\left(x - 4\right)}^{2} + {y}^{2} = 16$ touch the curves at the points $P$ and $Q$

Let, the equation of tangent be,

$y = mx + \dfrac{a}{m}$, where $a = 1$ is focus and $m$ is slope,

So, equation of tangent to parabola will be, $y = mx + \dfrac{1}{m}$

Also line is tangent to circle,

So, using perpendicular distance formula from centre $\left(4, 0\right)$to tangent which will be radius $r = 4$, we get,

\left|\dfrac{4m + \dfrac{1}{m}}{{m}^{2} + 1}\right| = 4$</p><p>$⇒ 16{m}^{2} + \dfrac{1}{{m}^{2}} + 8 = 16{m}^{2} + 16$</p><p>$⇒ {m}^{2} = \dfrac{1}{8}$</p><p>$⇒ m = \pm \dfrac{1}{2\sqrt{2}}$</p><p>Now if$m = \dfrac{1}{2\sqrt{2}}$then$P = \left(\dfrac{a}{{m}^{2}}, \dfrac{2a}{m}\right) = \left(8, 4\sqrt{2}\right)$</p><p>Now equation of tangent will be,$x - 2\sqrt{2}y + 8 = 0$</p><p>And$Q$will be foot of perpendicular from \left(4, 0\right)$ on tangent$x - 2\sqrt{2}y + 8 = 0$ ,

So, using foot of perpendicular formula we get,

$\dfrac{x - 4}{1} = \dfrac{y - 0}{- 2\sqrt{2}} = - \dfrac{4 - 0 + 8}{{1}^{2} + {\left(2\sqrt{2}\right)}^{2}}$

$⇒ \left(x, y\right) = \left(\dfrac{8}{3}, \dfrac{8\sqrt{2}}{3}\right)$

Hence, by distance formula we get,

${\left(PQ\right)}^{2} = {\left(\dfrac{16}{3}\right)}^{2} + {\left(\dfrac{4\sqrt{2}}{3}\right)}^{2}$

$⇒ {\left(PQ\right)}^{2} = \dfrac{256 + 32}{9} = 32$