If P is a point such that the ratio of the squares of the lengths of the tangents from P to the circles `x^(2)+y^(2)+2x-2y-20=0` and `x^(2)+y^(2)-4x+2y-44=0` is 2:3, then the locus of P is a circle with centre

Options

A. (7, -8)
B. (-7, 8)
C. (7, 8)
D. (-7, -8)

Answer

To solve the problem, we need to find the locus of point P such that the ratio of the squares of the lengths of the tangents from P to the given circles is 2:3.

### Step-by-Step Solution:

1. **Identify the circles**:
The equations of the circles are:
- Circle 1: \( x^2 + y^2 + 2x - 2y - 20 = 0 \)
- Circle 2: \( x^2 + y^2 - 4x + 2y - 44 = 0 \)

2. **Rearranging the circle equations**:
We can rewrite the equations in standard form:
- Circle 1:
\[ (x^2 + 2x + y^2 - 2y) = 20 \]
Completing the square:
\[ (x + 1)^2 + (y - 1)^2 = 22 \]
So, the center is \( (-1, 1) \) and radius \( \sqrt{22} \).

- Circle 2:
\[ (x^2 - 4x + y^2 + 2y) = 44 \]
Completing the square:
\[ (x - 2)^2 + (y + 1)^2 = 49 \]
So, the center is \( (2, -1) \) and radius \( 7 \).

3. **Length of the tangent from point P**:
The length of the tangent from a point \( P(x, y) \) to a circle \( (x - h)^2 + (y - k)^2 = r^2 \) is given by:
\[ L = \sqrt{(x - h)^2 + (y - k)^2 - r^2} \]
For Circle 1:
\[ L_1 = \sqrt{(x + 1)^2 + (y - 1)^2 - 22} \]
For Circle 2:
\[ L_2 = \sqrt{(x - 2)^2 + (y + 1)^2 - 49} \]

4. **Setting up the ratio**:
According to the problem, the ratio of the squares of the lengths of the tangents is given as:
\[ \frac{L_1^2}{L_2^2} = \frac{2}{3} \]
Therefore,
\[ \frac{(x + 1)^2 + (y - 1)^2 - 22}{(x - 2)^2 + (y + 1)^2 - 49} = \frac{2}{3} \]

5. **Cross-multiplying**:
Cross-multiplying gives:
\[ 3[(x + 1)^2 + (y - 1)^2 - 22] = 2[(x - 2)^2 + (y + 1)^2 - 49] \]

6. **Expanding both sides**:
Expanding the left side:
\[ 3[(x^2 + 2x + 1 + y^2 - 2y + 1 - 22)] = 3[x^2 + y^2 + 2x - 2y - 20] \]
Expanding the right side:
\[ 2[(x^2 - 4x + 4 + y^2 + 2y + 1 - 49)] = 2[x^2 + y^2 - 4x + 2y - 44] \]

7. **Combining like terms**:
After expanding and simplifying, we get:
\[ 3x^2 + 3y^2 + 6x - 12y - 60 = 2x^2 + 2y^2 - 8x + 4y - 88 \]
Rearranging gives:
\[ x^2 + y^2 + 14x - 16y + 28 = 0 \]

8. **Finding the center of the locus**:
The equation can be rewritten in the standard form:
\[ (x^2 + 14x) + (y^2 - 16y) + 28 = 0 \]
Completing the square:
\[ (x + 7)^2 - 49 + (y - 8)^2 - 64 + 28 = 0 \]
Simplifying gives:
\[ (x + 7)^2 + (y - 8)^2 = 85 \]
Thus, the center of the circle is \( (-7, 8) \).

### Final Answer:
The locus of point P is a circle with center at \( (-7, 8) \).