Circles Part 1

Equation of tangent to $y^{2}=30x$ is, $y=mx+\frac{30}{4m}$ Now, this tangent passes through $(-30,0)$. $\therefore 0=-30m+\frac{30}{4m}$ ⇒ $\frac{30}{4m}=30m$ ⇒ $m^{2}=\frac{1}{4}$ ⇒ $m=\pm\frac{1}{2}$ ∴ Equation of tangent is $y=\frac{x}{2}+15$ or$y=-\frac{x}{2}-15$ Now equation of circle is $x^{2}+y^{2}+30x+\frac{675}{4}=0$ Let perpendicular distance of the tangent from the centre $(-15,0)$ of the circle = p ∴ $p=\frac{\left|-\frac{15}{2}+15\right|}{\sqrt{1+\frac{1}{4}}}=3\sqrt{5}$ ∴ Length of chord $=2\sqrt{r^{2}-p^{2}}$ $=2\sqrt{\left(15^{2}+0-\frac{675}{4}\right)-45}=3\sqrt{5}$ where, r is radius of the given circle.