The point from which the tangents to the circles `x^2 +y^2-8x + 40 = 0,5x^2+5y^2 -25x +80=0,`and `x^2 +y^2-8x + 16y + 160 = 0` are equal in length, is

Options

A. (8, 15/2)
B. (-8, 15/2)
C. (8, -15/2)
D. none of these

Explanation

To find the point from which the tangents to the given circles are equal in length, we need to determine the radical center of the three circles. Here's a step-by-step solution:

### Step 1: Write the equations of the circles
The equations of the circles given are:
1. \( C_1: x^2 + y^2 - 8x + 40 = 0 \)
2. \( C_2: 5x^2 + 5y^2 - 25x + 80 = 0 \)
3. \( C_3: x^2 + y^2 - 8x + 16y + 160 = 0 \)

### Step 2: Convert the second circle to standard form
To make the coefficients of \(x^2\) and \(y^2\) equal to 1, we divide the entire equation of \(C_2\) by 5:
\[ C_2: x^2 + y^2 - 5x + 16 = 0 \]

### Step 3: Find the radical axis between the first two circles
To find the radical axis between \(C_1\) and \(C_2\), we equate their equations:
\[ C_1 = C_2 \]
This gives:
\[ x^2 + y^2 - 8x + 40 = x^2 + y^2 - 5x + 16 \]
Canceling \(x^2 + y^2\) from both sides:
\[ -8x + 40 = -5x + 16 \]
Rearranging gives:
\[ -8x + 5x = 16 - 40 \]
\[ -3x = -24 \implies x = 8 \]

### Step 4: Find the value of \(y\) using the first and third circles
Now, we equate \(C_1\) and \(C_3\):
\[ C_1 = C_3 \]
This gives:
\[ x^2 + y^2 - 8x + 40 = x^2 + y^2 - 8x + 16y + 160 \]
Canceling \(x^2 + y^2 - 8x\) from both sides:
\[ 40 = 16y + 160 \]
Rearranging gives:
\[ 40 - 160 = 16y \]
\[ -120 = 16y \implies y = -\frac{120}{16} = -\frac{15}{2} \]

### Step 5: Write the coordinates of the radical center
Thus, the coordinates of the point from which the tangents to the circles are equal in length are:
\[ (8, -\frac{15}{2}) \]

### Final Answer
The point is \( (8, -\frac{15}{2}) \).

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