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

Options

A. `2ax+2by-(a^(2)+b^(2)+p^(2))=0`
B. `2ax+2by-(a^(2)-b^(2)+p^(2))=0`
C. `x^(2)+y^(2)-3ax-4by+(a^(2)+b^(2)-p^(2))=0`
D. `x^(2)+y^(2)-2ax-3by+(a^(2)-b^(2)-p^(2))=0`

Answer

To find the equation of the locus of the center of a circle that passes through the point \((a, b)\) and cuts the circle given by the equation \(x^2 + y^2 = p^2\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the center of the circle**: Let the center of the required circle be \((\alpha, \beta)\).

2. **Write the equation of the circle**: The general equation of a circle with center \((\alpha, \beta)\) and radius \(r\) can be expressed as:
\[ x^2 + y^2 - 2\alpha x - 2\beta y + c_1 = 0 \]
where \(c_1\) is a constant related to the radius.

3. **Set up the condition for intersection**: Since the circle cuts the circle \(x^2 + y^2 = p^2\), we can equate the two equations:
\[ x^2 + y^2 - 2\alpha x - 2\beta y + c_1 - p^2 = 0 \]

4. **Substitute the point through which the circle passes**: The circle passes through the point \((a, b)\). Substituting \(x = a\) and \(y = b\) into the circle's equation gives:
\[ a^2 + b^2 - 2\alpha a - 2\beta b + c_1 = 0 \]

5. **Express \(c_1\)**: From the intersection condition, we can set \(c_1 = p^2\) (since the circle intersects at the radius \(p\)):
\[ c_1 - p^2 = 0 \implies c_1 = p^2 \]

6. **Substituting \(c_1\) back**: Now substituting \(c_1\) back into the equation gives:
\[ a^2 + b^2 - 2\alpha a - 2\beta b + p^2 = 0 \]

7. **Rearranging the equation**: Rearranging the above equation leads to:
\[ -2\alpha a - 2\beta b + a^2 + b^2 + p^2 = 0 \]

8. **Multiply through by -1**: To simplify, multiply the entire equation by -1:
\[ 2\alpha a + 2\beta b - a^2 - b^2 - p^2 = 0 \]

9. **Final form**: Rearranging gives us the required equation of the locus of the center \((\alpha, \beta)\):
\[ 2ax + 2by = a^2 + b^2 + p^2 \]

### Final Equation:
The equation of the locus of the center of the circle is:
\[ 2ax + 2by = a^2 + b^2 + p^2 \]