1. Write the circle in standard form

Given circle: $4x^2+4y^2+120x+675=0$ Divide by $4$: $x^2+y^2+30x+\frac{675}{4}=0$ Complete the square in $x$: $x^2+30x=(x+15)^2-225$ So, $(x+15)^2+y^2=225-\frac{675}{4}=\frac{225}{4}$ Hence the circle has

• centre $C(-15,0)$
• radius $r=\frac{15}{2}$.

2. Equation of tangent(s) from the point $(-30,0)$ to the parabola

Parabola: $y^2=30x$ This is of the form $y^2=4ax$ with $4a=30 \implies a=\frac{15}{2}$

A tangent to $y^2=4ax$ with slope $m$ is $y=mx+\frac{a}{m}$ So here, $y=mx+\frac{15/2}{m}=mx+\frac{15}{2m}$ Since it passes through $(-30,0)$, $0=m(-30)+\frac{15}{2m}$ $-30m+\frac{15}{2m}=0$ Multiply by $2m$: $-60m^2+15=0$ $m^2=\frac{1}{4}$ $m=\pm \frac12$

Thus the required line(s) are: $y=\frac12 x+15$ quad \text{and} \quad $y=-\frac12 x-15$ Both pass through $(-30,0)$.

3. Distance of either line from the centre of the circle

Take the line $y=\frac12 x+15$ Standard form: $x-2y+30=0$ Distance of centre $(-15,0)$ from this line: $d=\frac{|(-15)-2(0)+30|}{\sqrt{1^2+(-2)^2}}=\frac{15}{\sqrt5}=3\sqrt5$

4. Length of chord cut by a line at distance $d$ from centre

For a circle of radius $r$, chord length is $L=2\sqrt{r^2-d^2}$ Here, $r=\frac{15}{2}, \qquad d=3\sqrt5$ So, $L=2\sqrt{\left(\frac{15}{2}\right)^2-(3\sqrt5)^2}$ $=2\sqrt{\frac{225}{4}-45}$ $=2\sqrt{\frac{225-180}{4}}$ $=2\sqrt{\frac{45}{4}}$ $=\sqrt{45}=3\sqrt5$

5. Check both tangents

The other tangent $y=-\frac12 x-15$ is symmetric about the $x$-axis, and the centre lies on the $x$-axis, so its distance from the centre is the same. Hence the chord length is again the same.

Therefore, the chord length is: $\boxed{3\sqrt5}$

So the correct option is D.