If (h,k) is the centre of a circle passing through the origin then its equation is

Options

A. `x^(2) + y^(2) - hx - ky = 0 `
B. `x^(2) + y^(2) + hx - ky = 0 `
C. ` x^(2) + y^(2) + 2hx + 2ky = 0 `
D. `x^(2) + y^(2) - 2hx - 2ky = 0 `

Answer

To derive the equation of a circle with center (h, k) that passes through the origin (0, 0), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the center and radius**:
The center of the circle is given as (h, k). Since the circle passes through the origin (0, 0), we can find the radius (r) using the distance formula. The radius is the distance from the center (h, k) to the origin (0, 0).

\[ r = \sqrt{(h - 0)^2 + (k - 0)^2} = \sqrt{h^2 + k^2} \]

2. **Write the general equation of the circle**:
The general equation of a circle with center (h, k) and radius r is given by:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

Substituting the value of r we found:

\[ (x - h)^2 + (y - k)^2 = (\sqrt{h^2 + k^2})^2 \]

This simplifies to:

\[ (x - h)^2 + (y - k)^2 = h^2 + k^2 \]

3. **Expand the equation**:
Now, we expand the left side of the equation:

\[ (x - h)^2 = x^2 - 2hx + h^2 \]
\[ (y - k)^2 = y^2 - 2ky + k^2 \]

Combining these, we have:

\[ x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = h^2 + k^2 \]

4. **Simplify the equation**:
Now, we can simplify the equation by subtracting \(h^2 + k^2\) from both sides:

\[ x^2 - 2hx + h^2 + y^2 - 2ky + k^2 - h^2 - k^2 = 0 \]

This simplifies to:

\[ x^2 + y^2 - 2hx - 2ky = 0 \]

5. **Final equation**:
Thus, the equation of the circle with center (h, k) that passes through the origin is:

\[ x^2 + y^2 - 2hx - 2ky = 0 \]