A variable chord is drawn through the origin to the circle `x^(2)+y^(2) -2ax=0`. The locus of the centre of the circle drawn on this chord as diameter is

Options

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

Answer

To solve the problem, we need to find the locus of the center of the circle drawn on a variable chord through the origin to the circle given by the equation \( x^2 + y^2 - 2ax = 0 \).

### Step-by-Step Solution:

1. **Identify the Circle's Center and Radius**:
The given circle is \( x^2 + y^2 - 2ax = 0 \). We can rewrite it as:
\[ x^2 + y^2 = 2ax \]
The center of the circle can be found using the formula \((-g, -f)\) where \(g\) and \(f\) are the coefficients from the general circle equation \(x^2 + y^2 + 2gx + 2fy + c = 0\). Here, \(g = -a\) and \(f = 0\), so the center is \((a, 0)\) and the radius is \(a\).

2. **Define the Chord**:
Let the endpoints of the chord be \(B(b, c)\) and \(C(0, 0)\) (the origin). The midpoint \(M\) of the chord \(BC\) is given by:
\[ M\left(\frac{b + 0}{2}, \frac{c + 0}{2}\right) = \left(\frac{b}{2}, \frac{c}{2}\right) \]

3. **Express the Endpoints in Terms of the Midpoint**:
Let the coordinates of the midpoint be \(M(h, k)\). Therefore, we have:
\[ h = \frac{b}{2} \quad \text{and} \quad k = \frac{c}{2} \]
This implies:
\[ b = 2h \quad \text{and} \quad c = 2k \]

4. **Substitute into the Circle Equation**:
Since the endpoints \(B(b, c)\) lie on the circle, we substitute \(b\) and \(c\) into the circle's equation:
\[ b^2 + c^2 - 2ab = 0 \]
Substituting \(b = 2h\) and \(c = 2k\):
\[ (2h)^2 + (2k)^2 - 2a(2h) = 0 \]
Simplifying this gives:
\[ 4h^2 + 4k^2 - 4ah = 0 \]
Dividing through by 4:
\[ h^2 + k^2 - ah = 0 \]

5. **Rearranging the Equation**:
Rearranging the equation gives:
\[ h^2 + k^2 = ah \]
This represents the locus of the midpoint \(M(h, k)\) of the chord.

6. **Final Locus Equation**:
To express this in standard form, we can rewrite it as:
\[ h^2 - ah + k^2 = 0 \]
This is the equation of a circle with center \(\left(\frac{a}{2}, 0\right)\) and radius \(\frac{a^2}{4}\).

### Conclusion:
The locus of the center of the circle drawn on the chord as a diameter is given by the equation:
\[ x^2 + y^2 - ax = 0 \]