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-4y-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 the point P such that the ratio of the squares of the lengths of the tangents from P to the two given circles is 2:3. Let’s break it down step by step.

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

### Step 2: Rewrite the equations in standard form
We will convert both equations into standard form (i.e., \((x - h)^2 + (y - k)^2 = r^2\)).

**For the first circle:**
\[ x^2 + 2x + y^2 - 4y - 20 = 0 \]
Completing the square for \(x\) and \(y\):
\[ (x^2 + 2x + 1) + (y^2 - 4y + 4) = 25 \]
\[ (x + 1)^2 + (y - 2)^2 = 25 \]
Thus, the center is \((-1, 2)\) and the radius is \(5\).

**For the second circle:**
\[ x^2 - 4x + y^2 + 2y - 44 = 0 \]
Completing the square:
\[ (x^2 - 4x + 4) + (y^2 + 2y + 1) = 49 \]
\[ (x - 2)^2 + (y + 1)^2 = 49 \]
Thus, the center is \((2, -1)\) and the radius is \(7\).

### Step 3: Set up the ratio of the lengths of the tangents
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 the two circles, we denote the lengths of the tangents as \(L_1\) and \(L_2\).

The squares of the lengths of the tangents are:
\[ L_1^2 = (x + 1)^2 + (y - 2)^2 - 25 \]
\[ L_2^2 = (x - 2)^2 + (y + 1)^2 - 49 \]

### Step 4: Set up the equation based on the given ratio
According to the problem, the ratio of the squares of the lengths of the tangents is:
\[ \frac{L_1^2}{L_2^2} = \frac{2}{3} \]
This leads to:
\[ 3L_1^2 = 2L_2^2 \]

### Step 5: Substitute the expressions for \(L_1^2\) and \(L_2^2\)
Substituting the expressions we derived:
\[ 3\left((x + 1)^2 + (y - 2)^2 - 25\right) = 2\left((x - 2)^2 + (y + 1)^2 - 49\right) \]

### Step 6: Expand and simplify
Expanding both sides:
\[ 3((x + 1)^2 + (y - 2)^2) - 75 = 2((x - 2)^2 + (y + 1)^2) - 98 \]
Expanding gives:
\[ 3(x^2 + 2x + 1 + y^2 - 4y + 4) - 75 = 2(x^2 - 4x + 4 + y^2 + 2y + 1) - 98 \]
\[ 3x^2 + 6x + 3 + 3y^2 - 12y + 12 - 75 = 2x^2 - 8x + 8 + 2y^2 + 4y + 2 - 98 \]

### Step 7: Combine like terms
Combining like terms on both sides leads to:
\[ 3x^2 + 3y^2 + 6x - 12y - 60 = 2x^2 + 2y^2 - 8x + 4y - 88 \]
Rearranging gives:
\[ x^2 + y^2 + 14x - 16y + 28 = 0 \]

### Step 8: Rewrite in standard form
To find the center of the circle, we complete the square:
\[ (x^2 + 14x) + (y^2 - 16y) = -28 \]
Completing the square:
\[ (x + 7)^2 - 49 + (y - 8)^2 - 64 = -28 \]
This simplifies to:
\[ (x + 7)^2 + (y - 8)^2 = 85 \]
Thus, the center of the locus of point P is \((-7, 8)\).

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

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