The co-ordinates of the point from which the lengths of tangents to the three circles `x^(2)+y^(2)=1, x^(2)+y^(2)+8x+15=0, x^(2)+y^(2)+10y+24=0` are equal is

Options

A. (-2,1)
B. (1,-2)
C. (2,3)
D. `(-2, -5//2)`

Answer

To find the coordinates of the point from which the lengths of tangents to the three circles are equal, we will follow these steps:

### Step 1: Identify the equations of the circles
The equations of the three circles are:
1. \( C_1: x^2 + y^2 = 1 \)
2. \( C_2: x^2 + y^2 + 8x + 15 = 0 \)
3. \( C_3: x^2 + y^2 + 10y + 24 = 0 \)

### Step 2: Rewrite the equations in standard form
For \( C_2 \) and \( C_3 \), we can rewrite them in standard form by completing the square.

- For \( C_2 \):
\[ x^2 + 8x + y^2 + 15 = 0 \implies (x + 4)^2 + y^2 = 1 \]
This means the center is at \( (-4, 0) \) and the radius is \( 1 \).

- For \( C_3 \):
\[ x^2 + y^2 + 10y + 24 = 0 \implies x^2 + (y + 5)^2 = 1 \]
This means the center is at \( (0, -5) \) and the radius is \( 1 \).

### Step 3: Use the formula for the length of the tangent
The length of the tangent from a point \( (x_1, y_1) \) to a circle given by \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is given by:
\[ L = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c} \]

### Step 4: Calculate the lengths of tangents for each circle
1. For \( C_1 \):
\[ L_1 = \sqrt{x_1^2 + y_1^2 - 1} \]

2. For \( C_2 \) (where \( g = 4 \), \( f = 0 \), \( c = -15 \)):
\[ L_2 = \sqrt{x_1^2 + y_1^2 + 8x_1 + 15} \]

3. For \( C_3 \) (where \( g = 0 \), \( f = 5 \), \( c = -24 \)):
\[ L_3 = \sqrt{x_1^2 + y_1^2 + 10y_1 + 24} \]

### Step 5: Set the lengths equal
Since the lengths of the tangents are equal, we can set \( L_1 = L_2 \) and \( L_1 = L_3 \).

1. Setting \( L_1 = L_2 \):
\[ \sqrt{x_1^2 + y_1^2 - 1} = \sqrt{x_1^2 + y_1^2 + 8x_1 + 15} \]
Squaring both sides:
\[ x_1^2 + y_1^2 - 1 = x_1^2 + y_1^2 + 8x_1 + 15 \]
Simplifying gives:
\[ -1 = 8x_1 + 15 \implies 8x_1 = -16 \implies x_1 = -2 \]

2. Setting \( L_1 = L_3 \):
\[ \sqrt{x_1^2 + y_1^2 - 1} = \sqrt{x_1^2 + y_1^2 + 10y_1 + 24} \]
Squaring both sides:
\[ x_1^2 + y_1^2 - 1 = x_1^2 + y_1^2 + 10y_1 + 24 \]
Simplifying gives:
\[ -1 = 10y_1 + 24 \implies 10y_1 = -25 \implies y_1 = -\frac{5}{2} \]

### Step 6: Conclusion
The coordinates of the point from which the lengths of tangents to the three circles are equal is:
\[ \boxed{(-2, -\frac{5}{2})} \]