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Step-by-Step Solution
Step 1: Identify the Circle and Its Centre & Radius
The given circle is:
$$4x^2 + 4y^2 + 120x + 675 = 0.$$
Divide through by 4 to simplify:
$$x^2 + y^2 + 30x + \frac{675}{4} = 0.$$
Complete the square for the $x$ terms:
$$x^2 + 30x = \left(x^2 + 30x + 225\right) - 225 = (x + 15)^2 - 225.$$
So the circle equation becomes:
$$(x+15)^2 + y^2 = 225 - \frac{675}{4} = \frac{900}{4} - \frac{675}{4} = \frac{225}{4}.$$
Hence, the centre of the circle is $(-15, 0)$ and its radius $r$ is
$$r = \sqrt{\frac{225}{4}} = \frac{15}{2}.$$
Step 2: Write the General Form of Tangent to the Parabola and Impose Point Condition
The parabola is
$$y^2 = 30x.$$
In the slope form, a tangent to $y^2 = 4ax$ is given by
$$y = mx + \frac{a}{m},$$
where $4a = 30 \implies a = \frac{30}{4} = 7.5.$
Thus, the tangent can be written as:
$$y = mx + \frac{7.5}{m} \quad \text{or equivalently} \quad y = mx + \frac{30}{4m}.$$
Because the tangent passes through the point $(-30, 0)$, substitute $x = -30$ and $y = 0$ into the line:
$$0 = m(-30) + \frac{30}{4m} \quad \Longrightarrow \quad -30m + \frac{30}{4m} = 0.$$
Multiply both sides by $4m$:
$$-120m^2 + 30 = 0 \quad \Longrightarrow \quad 120m^2 = 30 \quad \Longrightarrow \quad m^2 = \frac{30}{120} = \frac{1}{4}.$$
Thus,
$$m = \frac{1}{2} \quad \text{or} \quad m = -\frac{1}{2}.$$
Step 3: Choose an Appropriate Slope and Form the Line
Taking $m = \frac{1}{2}$, the tangent becomes:
$$y = \frac{x}{2} + 15.$$
Rewriting in standard form:
$$x - 2y + 30 = 0.$$
This line (chord of the given circle) must lie along $x - 2y + 30 = 0$.
Step 4: Find the Perpendicular Distance from Centre to the Line
To find the chord length, we calculate the perpendicular distance $P$ from the centre of the circle $(-15,0)$ to the line $x - 2y + 30 = 0.$
The formula for the perpendicular distance from $(x_1, y_1)$ to $Ax + By + C = 0$ is
$$P = \frac{|A x_1 + B y_1 + C|}{\sqrt{A^2 + B^2}}.$$
Here, $A=1$, $B=-2$, $C=30$, and $(x_1, y_1)=(-15,0).$ So,
$$P = \frac{|1 \cdot (-15) + (-2)\cdot 0 + 30|}{\sqrt{1^2 + (-2)^2}}
= \frac{|-15 + 30|}{\sqrt{1 + 4}}
= \frac{15}{\sqrt{5}}
= 3\sqrt{5}.$$
Step 5: Use the Chord-Length Formula
If $r$ is the circle’s radius and $P$ the perpendicular distance from the circle’s centre to the chord, then the chord length $l_{\text{AB}}$ is
$$l_{\text{AB}} = 2\sqrt{r^2 - P^2}.$$
We have $r = \frac{15}{2}$, hence $r^2 = \frac{225}{4},$ and $P=3\sqrt{5},$ so $P^2 = 9 \times 5 = 45.$
Thus,
$$r^2 - P^2 = \frac{225}{4} - 45 = \frac{225}{4} - \frac{180}{4} = \frac{45}{4}.$$
Hence,
$$l_{\text{AB}} = 2 \sqrt{\frac{45}{4}}
= 2 \cdot \frac{\sqrt{45}}{2}
= \sqrt{45}
= 3\sqrt{5}.$$
Final Answer:
The length of the chord is
$$3\sqrt{5}.$$
