The locus of the mid-points of the chords of the circle `x^(2)+y^(2)-2ax-2by=0` which subtend a right angle at the centre is

Options

A. `ax+by=0`
B. `ax+by=a^(2)+b^(2)`
C. `x^(2)+y^(2) =a^(2)+b^(2)`
D. none of these

Answer

To find the locus of the midpoints of the chords of the circle \(x^2 + y^2 - 2ax - 2by = 0\) that subtend a right angle at the center, we can follow these steps:

### Step 1: Identify the center and radius of the circle
The given equation of the circle can be rewritten in standard form. Completing the square for \(x\) and \(y\):

\[ x^2 - 2ax + y^2 - 2by = 0 \]

This can be rearranged to:

\[ (x - a)^2 + (y - b)^2 = a^2 + b^2 \]

Thus, the center \(C\) of the circle is at \((a, b)\) and the radius \(r\) is \(\sqrt{a^2 + b^2}\).

### Step 2: Consider a chord that subtends a right angle at the center
Let \(P(h, k)\) be the midpoint of the chord \(AB\) that subtends a right angle at the center \(C(a, b)\). According to the property of circles, if a chord subtends a right angle at the center, then the midpoint of that chord lies on a circle whose center is the center of the original circle and whose radius is \(\frac{r}{\sqrt{2}}\).

### Step 3: Calculate the distance from the center to the midpoint
The distance \(CP\) from the center \(C(a, b)\) to the midpoint \(P(h, k)\) can be expressed using the distance formula:

\[ CP = \sqrt{(h - a)^2 + (k - b)^2} \]

### Step 4: Set up the equation for the locus
Since \(CP\) is equal to \(\frac{r}{\sqrt{2}}\), we have:

\[ \sqrt{(h - a)^2 + (k - b)^2} = \frac{1}{\sqrt{2}} \sqrt{a^2 + b^2} \]

Squaring both sides gives:

\[ (h - a)^2 + (k - b)^2 = \frac{1}{2}(a^2 + b^2) \]

### Step 5: Rearranging the equation
Expanding the left side:

\[ (h^2 - 2ah + a^2) + (k^2 - 2bk + b^2) = \frac{1}{2}(a^2 + b^2) \]

Combining terms:

\[ h^2 + k^2 - 2ah - 2bk + a^2 + b^2 = \frac{1}{2}(a^2 + b^2) \]

Rearranging gives:

\[ h^2 + k^2 - 2ah - 2bk + \frac{1}{2}(a^2 + b^2) = 0 \]

### Step 6: Final form of the equation
This can be rewritten as:

\[ h^2 + k^2 - 2ah - 2bk = -\frac{1}{2}(a^2 + b^2) \]

This represents a circle with center \((a, b)\) and radius \(\sqrt{\frac{1}{2}(a^2 + b^2)}\).

### Conclusion
The locus of the midpoints of the chords of the circle that subtend a right angle at the center is a circle given by the equation:

\[ x^2 + y^2 = a^2 + b^2 \]