The equation of the circle passing through (1,2) and the points of intersection of the circles `x^2+y^2-8x-6y+21=0` and `x^2+y^2-2x-15=0` is

Options

A. `x^2+y^2+6x-2y+9=0`
B. `x^2+y^2-6x-4y+9=0`
C. `x^2+y^2-6x-4y+9=0`
D. `x^2+y^2-6x+4y+9=0`

Answer

To find the equation of the circle that passes through the point (1, 2) and the points of intersection of the circles given by the equations \(x^2 + y^2 - 8x - 6y + 21 = 0\) and \(x^2 + y^2 - 2x - 15 = 0\), we can follow these steps:

### Step 1: Identify the equations of the given circles
The equations of the circles are:
1. Circle 1: \(x^2 + y^2 - 8x - 6y + 21 = 0\)
2. Circle 2: \(x^2 + y^2 - 2x - 15 = 0\)

### Step 2: Use the family of circles that pass through the intersection of two circles
The general equation of the family of circles passing through the intersection of two circles can be expressed as:
\[ s_1 + \lambda s_2 = 0 \]
where \(s_1\) and \(s_2\) are the equations of the two circles, and \(\lambda\) is a parameter.

Substituting the equations of the circles, we have:
\[ (x^2 + y^2 - 8x - 6y + 21) + \lambda (x^2 + y^2 - 2x - 15) = 0 \]

### Step 3: Combine the equations
Combining the equations, we get:
\[ (1 + \lambda)x^2 + (1 + \lambda)y^2 - (8 + 2\lambda)x - (6 + 15\lambda)y + (21 - 15\lambda) = 0 \]

### Step 4: Substitute the point (1, 2)
Since the circle passes through the point (1, 2), we substitute \(x = 1\) and \(y = 2\) into the equation:
\[ (1 + \lambda)(1^2) + (1 + \lambda)(2^2) - (8 + 2\lambda)(1) - (6 + 15\lambda)(2) + (21 - 15\lambda) = 0 \]

Calculating this step-by-step:
1. \(1 + \lambda\)
2. \(1 + \lambda\) multiplied by \(4\) (since \(2^2 = 4\)) gives \(4 + 4\lambda\)
3. \(- (8 + 2\lambda)\) gives \(-8 - 2\lambda\)
4. \(- (6 + 15\lambda)(2)\) gives \(-12 - 30\lambda\)
5. \(21 - 15\lambda\)

Combining these:
\[ (1 + \lambda + 4 + 4\lambda - 8 - 2\lambda - 12 - 30\lambda + 21 - 15\lambda) = 0 \]
This simplifies to:
\[ (1 + 4 - 8 - 12 + 21) + (1 + 4 - 2 - 30 - 15)\lambda = 0 \]
Calculating the constant terms:
\[ 6 + (-42\lambda) = 0 \]
This leads to:
\[ -42\lambda = -6 \implies \lambda = \frac{1}{7} \]

### Step 5: Substitute \(\lambda\) back into the equation
Now substituting \(\lambda = \frac{1}{7}\) back into the combined equation:
\[ (1 + \frac{1}{7})x^2 + (1 + \frac{1}{7})y^2 - (8 + 2 \cdot \frac{1}{7})x - (6 + 15 \cdot \frac{1}{7})y + (21 - 15 \cdot \frac{1}{7}) = 0 \]
Calculating each term:
1. \(1 + \frac{1}{7} = \frac{8}{7}\)
2. \(- (8 + \frac{2}{7}) = - \frac{58}{7}\)
3. \(- (6 + \frac{15}{7}) = - \frac{57}{7}\)
4. \(21 - \frac{15}{7} = \frac{132}{7}\)

### Step 6: Final equation
Combining these gives:
\[ \frac{8}{7}x^2 + \frac{8}{7}y^2 - \frac{58}{7}x - \frac{57}{7}y + \frac{132}{7} = 0 \]
Multiplying through by \(7\) to eliminate the fraction:
\[ 8x^2 + 8y^2 - 58x - 57y + 132 = 0 \]
Dividing through by \(8\):
\[ x^2 + y^2 - \frac{29}{4}x - \frac{57}{8}y + \frac{33}{2} = 0 \]

This gives us the final equation of the required circle.