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. `(-8,(-15)/2)`

Answer

To solve the problem of finding the point from which the tangents to the given circles are equal in length, we will follow these steps:

### Step 1: Write down the equations of the circles
The equations of the circles given in the problem are:
1. \( S_1: x^2 + y^2 - 8x + 40 = 0 \)
2. \( S_2: 5x^2 + 5y^2 - 25x + 80 = 0 \)
3. \( S_3: x^2 + y^2 - 8x + 16y + 160 = 0 \)

### Step 2: Simplify the second circle's equation
To simplify \( S_2 \), we divide the entire equation by 5:
\[ S_2: x^2 + y^2 - 5x + 16 = 0 \]

### Step 3: Find the difference between the second and first circle's equations
We calculate \( S_2 - S_1 \):
\[ (x^2 + y^2 - 5x + 16) - (x^2 + y^2 - 8x + 40) = 0 \]
This simplifies to:
\[ -5x + 8x + 16 - 40 = 0 \implies 3x - 24 = 0 \]
Solving for \( x \):
\[ 3x = 24 \implies x = 8 \]

### Step 4: Find the difference between the third and first circle's equations
Next, we calculate \( S_3 - S_1 \):
\[ (x^2 + y^2 - 8x + 16y + 160) - (x^2 + y^2 - 8x + 40) = 0 \]
This simplifies to:
\[ 16y + 160 - 40 = 0 \implies 16y + 120 = 0 \]
Solving for \( y \):
\[ 16y = -120 \implies y = -\frac{120}{16} = -\frac{15}{2} \]

### Step 5: Combine the values of \( x \) and \( y \)
The point from which the tangents to the circles are equal in length is:
\[ (x, y) = \left(8, -\frac{15}{2}\right) \]

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

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