The locus of the points from which the lengths of the tangents to the two circles `x^(2)+y^(2)+4x+3=0, x^(2)+y^(2)-6x+5=0` are in the ratio 2:3 is a circle with centre

Options

A. (6,0)
B. (-6,0)
C. (0,6)
D. (0,-6)

Answer

To solve the problem, we need to find the locus of points from which the lengths of the tangents to the two given circles are in the ratio 2:3. Let's break down the solution step by step.

### Step 1: Write the equations of the circles in standard form

The given equations of the circles are:
1. \( x^2 + y^2 + 4x + 3 = 0 \)
2. \( x^2 + y^2 - 6x + 5 = 0 \)

We can rewrite these equations by completing the square.

For the first circle:
\[ x^2 + 4x + y^2 + 3 = 0 \implies (x + 2)^2 + y^2 = 1 \]
This circle has center \((-2, 0)\) and radius \(1\).

For the second circle:
\[ x^2 - 6x + y^2 + 5 = 0 \implies (x - 3)^2 + y^2 = 4 \]
This circle has center \((3, 0)\) and radius \(2\).

### Step 2: Length of tangents from an external point

Let \( P(h, k) \) be an external point. The lengths of the tangents from point \( P \) to the circles can be calculated using the formula:
\[ L = \sqrt{(h - x_0)^2 + (k - y_0)^2 - r^2} \]
where \((x_0, y_0)\) is the center of the circle and \(r\) is the radius.

For the first circle:
\[ L_1 = \sqrt{(h + 2)^2 + k^2 - 1} \]

For the second circle:
\[ L_2 = \sqrt{(h - 3)^2 + k^2 - 4} \]

### Step 3: Set up the ratio of the lengths

According to the problem, the ratio of the lengths of the tangents is given as:
\[ \frac{L_1}{L_2} = \frac{2}{3} \]

This can be rewritten as:
\[ 3L_1 = 2L_2 \]

### Step 4: Substitute the expressions for \(L_1\) and \(L_2\)

Substituting the expressions we derived:
\[ 3\sqrt{(h + 2)^2 + k^2 - 1} = 2\sqrt{(h - 3)^2 + k^2 - 4} \]

### Step 5: Square both sides to eliminate the square roots

Squaring both sides gives:
\[ 9((h + 2)^2 + k^2 - 1) = 4((h - 3)^2 + k^2 - 4) \]

### Step 6: Expand and simplify

Expanding both sides:
\[ 9(h^2 + 4h + 4 + k^2 - 1) = 4(h^2 - 6h + 9 + k^2 - 4) \]
\[ 9h^2 + 36h + 27 + 9k^2 - 9 = 4h^2 - 24h + 20 + 4k^2 \]

Combining like terms:
\[ 5h^2 + 5k^2 + 60h + 7 = 0 \]

### Step 7: Rearranging to find the center

Dividing the entire equation by 5:
\[ h^2 + k^2 + 12h + \frac{7}{5} = 0 \]

Completing the square for \(h\):
\[ (h + 6)^2 + k^2 = 36 - \frac{7}{5} \]
\[ (h + 6)^2 + k^2 = \frac{180 - 7}{5} = \frac{173}{5} \]

### Conclusion: Identify the center of the circle

From the equation \((h + 6)^2 + k^2 = \frac{173}{5}\), we can see that the center of the circle is at:
\[ (-6, 0) \]

### Final Answer:
The center of the circle is \((-6, 0)\).