The centres of the circles `x^(2)+y^(2)-6x-8y-7=0` and `x^(2)+y^(2)-4x-10y-3=0` are the ends of the diameter of the circles

Options

A. `x^(2)+y^(2)-5x-9y+26=0`
B. `x^(2)+y^(2)+5x-9y+14=0`
C. `x^(2)+y^(2)+5x-y-14=0`
D. `x^(2)+y^(2)+5x+y+14=0`

Answer

To find the equation of the circle whose diameter is defined by the centers of the given circles, we will follow these steps:

### Step 1: Identify the centers of the given circles
The equations of the circles are:
1. \( x^2 + y^2 - 6x - 8y - 7 = 0 \)
2. \( x^2 + y^2 - 4x - 10y - 3 = 0 \)

We can rewrite the general equation of a circle in the form:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

From the first circle, we have:
- \( 2g = -6 \) → \( g = -3 \)
- \( 2f = -8 \) → \( f = -4 \)
- The center of the first circle is \( (-g, -f) = (3, 4) \).

From the second circle, we have:
- \( 2g = -4 \) → \( g = -2 \)
- \( 2f = -10 \) → \( f = -5 \)
- The center of the second circle is \( (-g, -f) = (2, 5) \).

### Step 2: Find the equation of the circle with the diameter defined by the centers
The centers of the two circles are \( (3, 4) \) and \( (2, 5) \). The equation of a circle with these two points as the endpoints of the diameter can be expressed as:
\[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \]
where \( (x_1, y_1) = (3, 4) \) and \( (x_2, y_2) = (2, 5) \).

Substituting the values:
\[ (x - 3)(x - 2) + (y - 4)(y - 5) = 0 \]

### Step 3: Expand the equation
Expanding the equation:
1. For the \( x \) terms:
\[ (x - 3)(x - 2) = x^2 - 2x - 3x + 6 = x^2 - 5x + 6 \]
2. For the \( y \) terms:
\[ (y - 4)(y - 5) = y^2 - 5y - 4y + 20 = y^2 - 9y + 20 \]

Now, combine both expansions:
\[ x^2 - 5x + 6 + y^2 - 9y + 20 = 0 \]

### Step 4: Combine like terms
Combining the terms gives:
\[ x^2 + y^2 - 5x - 9y + 26 = 0 \]

### Final Equation
Thus, the equation of the circle whose diameter is defined by the centers of the given circles is:
\[ x^2 + y^2 - 5x - 9y + 26 = 0 \]