From origin chords are drawn to the cirlce `x^(2)+y^(2)-2px=0` then locus of midpoints of all such chords is

Options

A. `x^(2)+y^(2)-px=0`
B. `x^(2)+y^(2)+2px=0`
C. `x^(2)+y^(2)+px=0`
D. Does not exists

Answer

To find the locus of midpoints of all chords drawn from the origin to the circle given by the equation \(x^2 + y^2 - 2px = 0\), we can follow these steps:

### Step 1: Understand the Circle Equation
The given equation of the circle is:
\[ x^2 + y^2 - 2px = 0 \]
This can be rewritten as:
\[ x^2 + y^2 = 2px \]
This represents a circle centered at \((p, 0)\) with a radius of \(p\).

### Step 2: Parametrize the Chords
Since we are drawing chords from the origin \((0, 0)\) to the circle, we can express the line through the origin in slope-intercept form:
\[ y = mx \]
where \(m\) is the slope of the line.

### Step 3: Substitute into the Circle Equation
Substituting \(y = mx\) into the circle's equation:
\[ x^2 + (mx)^2 = 2px \]
This simplifies to:
\[ x^2 + m^2x^2 = 2px \]
or
\[ (1 + m^2)x^2 - 2px = 0 \]

### Step 4: Factor the Equation
Factoring out \(x\) (since \(x \neq 0\) for the chord):
\[ x((1 + m^2)x - 2p) = 0 \]
This gives us:
\[ (1 + m^2)x - 2p = 0 \]
Solving for \(x\):
\[ x = \frac{2p}{1 + m^2} \]

### Step 5: Find the Corresponding y-coordinate
Using \(y = mx\):
\[ y = m \left(\frac{2p}{1 + m^2}\right) = \frac{2pm}{1 + m^2} \]

### Step 6: Midpoint of the Chord
The coordinates of the point on the circle (other than the origin) are:
\[ \left(\frac{2p}{1 + m^2}, \frac{2pm}{1 + m^2}\right) \]
The midpoint \(H, K\) of the chord from the origin to this point is given by:
\[ H = \frac{0 + \frac{2p}{1 + m^2}}{2} = \frac{p}{1 + m^2} \]
\[ K = \frac{0 + \frac{2pm}{1 + m^2}}{2} = \frac{pm}{1 + m^2} \]

### Step 7: Relate H and K
Now we have:
\[ H = \frac{p}{1 + m^2}, \quad K = \frac{pm}{1 + m^2} \]
From these, we can express \(m\) in terms of \(H\) and \(K\):
\[ m = \frac{K(1 + m^2)}{p} \]
Substituting \(H\) into this gives:
\[ K = \frac{p \cdot m}{1 + m^2} \implies K(1 + m^2) = pm \]

### Step 8: Eliminate m
Squaring both sides:
\[ K^2(1 + m^2)^2 = p^2m^2 \]
Expanding and rearranging leads to:
\[ K^2 + K^2m^2 = p^2m^2 \]
This can be rearranged to form:
\[ K^2 = (p^2 - K^2)m^2 \]
This implies:
\[ \frac{K^2}{p^2 - K^2} = m^2 \]

### Step 9: Locus Equation
Finally, substituting back, we find:
\[ H^2 + K^2 = \frac{p^2}{1 + m^2} \]
Replacing \(H\) and \(K\) with \(x\) and \(y\):
\[ x^2 + y^2 = px \]
Thus, the locus of midpoints of all such chords is given by:
\[ x^2 + y^2 - px = 0 \]

### Final Answer
The locus of midpoints of all such chords is:
\[ x^2 + y^2 - px = 0 \]