A Circle centre=(-15,0) and radius=`15/2`. Chord to circle passes through `(-30,0)` & tangent to `y^2=30x`. Length of chord=

Answer

To solve the problem, we need to find the length of the chord of a circle that passes through the point (-30, 0) and is tangent to the parabola given by the equation \(y^2 = 30x\). The center of the circle is at (-15, 0) and the radius is \( \frac{15}{2} \).

### Step-by-Step Solution:

1. **Identify the Circle's Equation**:
The equation of the circle can be expressed as:
\[ (x + 15)^2 + y^2 = \left(\frac{15}{2}\right)^2 \]
Simplifying this gives:
\[ (x + 15)^2 + y^2 = \frac{225}{4} \]

2. **Find the Tangent to the Parabola**:
The equation of the parabola is \(y^2 = 30x\). We can rewrite this as:
\[ y^2 = 4px \quad \text{where } p = \frac{30}{4} = 7.5 \]
The slope of the tangent line to the parabola at any point can be expressed as \(y = mx + \frac{p}{m}\).

3. **Determine the Slope of the Tangent**:
We need to find the slope \(m\) such that the tangent line passes through the point (-30, 0). Substituting \(x = -30\) and \(y = 0\) into the tangent equation gives:
\[ 0 = m(-30) + \frac{7.5}{m} \]
Rearranging this leads to:
\[ 30m^2 + 7.5 = 0 \implies m^2 = -\frac{7.5}{30} \implies m^2 = -0.25 \]
This indicates that the tangent slopes are \(m = \pm \frac{1}{2}\).

4. **Equation of the Tangent Line**:
Using \(m = \frac{1}{2}\), the equation of the tangent line becomes:
\[ y = \frac{1}{2}x + \frac{15}{2} \]
Rearranging gives:
\[ x - 2y + 15 = 0 \]

5. **Calculate the Perpendicular Distance from the Circle's Center to the Tangent**:
The center of the circle is at (-15, 0). The formula for the perpendicular distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Substituting \(A = 1\), \(B = -2\), \(C = 15\), and \((x_0, y_0) = (-15, 0)\):
\[ d = \frac{|1(-15) - 2(0) + 15|}{\sqrt{1^2 + (-2)^2}} = \frac{|0|}{\sqrt{5}} = 0 \]

6. **Using Pythagorean Theorem**:
The radius of the circle is \(r = \frac{15}{2}\). The length of the chord \(AB\) can be found using the relationship:
\[ AO^2 = AC^2 + OC^2 \]
Here, \(AO = r\), \(OC = d\), and \(AC\) is half the length of the chord:
\[ \left(\frac{15}{2}\right)^2 = AC^2 + d^2 \]
Since \(d = 0\):
\[ AC^2 = \left(\frac{15}{2}\right)^2 = \frac{225}{4} \]
Therefore, \(AC = \frac{15}{2}\).

7. **Finding the Length of the Chord**:
The length of the chord \(AB\) is:
\[ AB = 2 \times AC = 2 \times \frac{15}{2} = 15 \]

### Final Answer:
The length of the chord is \(15\).