Let `C_1 and C_2` are circles defined by `x^2+y^2 -20x+64=0` and `x^2+y^2+30x +144=0`. The length of the shortest line segment PQ that is tangent to `C_1` at P and to `C_2` at Q is

Options

A. 15
B. 18
C. 20
D. 24

Explanation

To find the length of the shortest line segment \( PQ \) that is tangent to the circles \( C_1 \) and \( C_2 \) at points \( P \) and \( Q \) respectively, we will follow these steps:

### Step 1: Identify the equations of the circles
The equations of the circles are given as:
1. \( C_1: x^2 + y^2 - 20x + 64 = 0 \)
2. \( C_2: x^2 + y^2 + 30x + 144 = 0 \)

### Step 2: Rewrite the equations in standard form
To find the centers and radii, we need to rewrite these equations in the standard form \( (x - h)^2 + (y - k)^2 = r^2 \).

For \( C_1 \):
\[ x^2 - 20x + y^2 + 64 = 0 \implies (x^2 - 20x + 100) + y^2 = 36 \implies (x - 10)^2 + y^2 = 6^2 \]
Thus, the center \( C_1 \) is \( (10, 0) \) and the radius \( r_1 = 6 \).

For \( C_2 \):
\[ x^2 + 30x + y^2 + 144 = 0 \implies (x^2 + 30x + 225) + y^2 = 81 \implies (x + 15)^2 + y^2 = 9^2 \]
Thus, the center \( C_2 \) is \( (-15, 0) \) and the radius \( r_2 = 9 \).

### Step 3: Calculate the distance \( d \) between the centers of the circles
The distance \( d \) between the centers \( (10, 0) \) and \( (-15, 0) \) is given by:
\[ d = \sqrt{(10 - (-15))^2 + (0 - 0)^2} = \sqrt{(10 + 15)^2} = \sqrt{25^2} = 25 \]

### Step 4: Use the formula for the length of the transverse tangent
The length \( L \) of the transverse tangent between two circles is given by:
\[ L = \sqrt{d^2 - (r_1 + r_2)^2} \]
Substituting the values we found:
\[ L = \sqrt{25^2 - (6 + 9)^2} = \sqrt{625 - 15^2} = \sqrt{625 - 225} = \sqrt{400} \]

### Step 5: Calculate the length
\[ L = \sqrt{400} = 20 \]

### Conclusion
The length of the shortest line segment \( PQ \) that is tangent to \( C_1 \) at \( P \) and to \( C_2 \) at \( Q \) is \( 20 \).

---