Locus of the mid points of the chords of the circle `x^2+y^2=a^2` which pass through the fixed point `(h ,k)` is `x^2+y^2+2h x+2k y=0` `x^2+y^2-2h x-2k y=0` `x^2+y^2+h x+k y=0` `x^2+y^2-h x-k y=0` `x^2+y^2+h x-k y=0`

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 find the locus of the midpoints of the chords of the circle \(x^2 + y^2 = a^2\) that pass through the fixed point \((h, k)\), we can follow these steps:

### Step 1: Define the midpoint
Let the coordinates of the midpoint of the chord be \((l, m)\).

### Step 2: Use the equation of the chord
The equation of the chord can be expressed in the form:
\[ lx + my = a^2 \]
This represents the equation of a chord of the circle where \((l, m)\) is the midpoint.

### Step 3: Relate the midpoint to the circle
Since \((l, m)\) lies on the circle, we can also write:
\[ l^2 + m^2 = a^2 \]

### Step 4: Equate the two equations
From the chord equation, we have:
\[ lx + my - a^2 = 0 \]
And from the circle, we have:
\[ l^2 + m^2 - a^2 = 0 \]
Now, substituting the fixed point \((h, k)\) into the chord equation gives:
\[ lh + mk = a^2 \]

### Step 5: Substitute \(l\) and \(m\)
Replace \(l\) with \(x\) and \(m\) with \(y\) in the equation:
\[ xh + yk = x^2 + y^2 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ x^2 + y^2 - hx - ky = 0 \]

### Final Equation
Thus, the required locus of the midpoints of the chords of the circle that pass through the fixed point \((h, k)\) is:
\[ x^2 + y^2 - hx - ky = 0 \]

### Conclusion
The correct option from the given choices is:
\[ x^2 + y^2 - hx - ky = 0 \]