If a circle passes through the point (a,b) and cuts the circle `x^2+y^2=k^2` orthogonally then the locus of its centre is

Options

A. `2ax+2by=a^2+b^2+k^2`
B. `ax+by=a^2+b^2+k^2`
C. `x^2+y^2+2ax+2by+k^2=0`
D. `x^2+y^2-2ax+2by+k^2=0`

Answer

To solve the problem, we need to find the locus of the center of a circle that passes through the point (a, b) and intersects the circle given by the equation \(x^2 + y^2 = k^2\) orthogonally.

### Step-by-Step Solution:

1. **Understanding the Circles**:
- Let the first circle (which we will denote as \(C_1\)) have the general equation:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
Here, the center of this circle is \((-g, -f)\) and the radius \(r_1\) can be expressed as:
\[ r_1 = \sqrt{g^2 + f^2 - c} \]
- The second circle (denoted as \(C_2\)) is given by:
\[ x^2 + y^2 = k^2 \]
which has its center at the origin (0, 0) and radius \(r_2 = k\).

2. **Condition for Orthogonality**:
- For two circles to intersect orthogonally, the following condition must hold:
\[ 2(g_1 g_2 + f_1 f_2) = c_1 + c_2 \]
- In our case:
- For \(C_1\): \(g_1 = g\), \(f_1 = f\), \(c_1 = c\)
- For \(C_2\): \(g_2 = 0\), \(f_2 = 0\), \(c_2 = -k^2\)
- Substituting these values into the orthogonality condition gives:
\[ 2(g \cdot 0 + f \cdot 0) = c - k^2 \]
Simplifying this, we find:
\[ 0 = c - k^2 \implies c = k^2 \]

3. **Substituting \(c\) back into the Circle Equation**:
- Now we can write the equation of the first circle as:
\[ x^2 + y^2 + 2gx + 2fy + k^2 = 0 \]

4. **Condition for the Circle to Pass through (a, b)**:
- Since the circle passes through the point \((a, b)\), we substitute \(x = a\) and \(y = b\) into the circle equation:
\[ a^2 + b^2 + 2ga + 2fb + k^2 = 0 \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ 2ga + 2fb = - (a^2 + b^2 + k^2) \]
- Dividing through by 2:
\[ ga + fb = -\frac{1}{2}(a^2 + b^2 + k^2) \]

6. **Expressing in Terms of the Center**:
- Recall that \(g = -x\) and \(f = -y\) (since the center of the circle is \((-g, -f)\)):
\[ -xa - yb = -\frac{1}{2}(a^2 + b^2 + k^2) \]
- This simplifies to:
\[ xa + yb = \frac{1}{2}(a^2 + b^2 + k^2) \]

7. **Final Locus Equation**:
- The final equation represents the locus of the center of the circle:
\[ ax + by = \frac{1}{2}(a^2 + b^2 + k^2) \]

### Conclusion:
The locus of the center of the circle that passes through the point \((a, b)\) and intersects the circle \(x^2 + y^2 = k^2\) orthogonally is given by:
\[ ax + by = \frac{1}{2}(a^2 + b^2 + k^2) \]