If a line along a chord of the circle 4 x ^ 2 + 4 y ^ 2 + 120x +...

We have,

$4{x}^{2} + 4{y}^{2} + 120x + 675 = 0$

$⇒ {x}^{2} + {y}^{2} + 30x + \dfrac{675}{4} = 0$

Centre$≡$\left(- 15, 0\right)$and radius= \sqrt{225 - \dfrac{675}{4}} =$\dfrac{15}{2}$units</p><p>Let the$chord cuts the circle at A \text{ and } B$and passes through$P\left(- 30, 0\right)

Equation of tangent to parabola ${y}^{2} = 30x$ is

$y = mx +$\dfrac{30}{4m}$. . . \left(i\right)$

It passes through \left(- 30, 0\right)$, so</span></p><p><span style="color:null;">$0 = - 30m + \dfrac{30}{4m}$</span></p><p><span style="color:null;">$⇒ 4{m}^{2} = 1$</span></p><p><span style="color:null;">$⇒ m = \pm \dfrac{1}{2}$</span></p><p><span style="color:null;">For$m = \dfrac{1}{2}$, we have</span></p><p><span style="color:null;">$y = \dfrac{x}{2} + 15$</span></p><p><span style="color:null;">$⇒ x - 2y + 30 = 0$</span></p><p><span style="color:null;">Length of$OB = \dfrac{- 15 + 0 + 30}{\sqrt{5}} = 3\sqrt{5}$</span></p><p><span style="color:null;">Length of$BC = \sqrt{\dfrac{225}{4} - 45} ⇒ BC = \dfrac{3\sqrt{5}}{2}$</span></p><p><span style="color:null;">Length of chord$AC = 2 \times \dfrac{3\sqrt{5}}{2} = 3\sqrt{5}$.</span></p><p><span style="color:null;">Similarly for$m = - \dfrac{1}{2}$, length of chord$AC = 3\sqrt{5}.