Find the locus of the mid point of the circle `x^2+y^2=a^2` which subtend a right angle at the point (p,q)

Answer

To find the locus of the midpoint of the chord of the circle \(x^2 + y^2 = a^2\) that subtends a right angle at the point \((p, q)\), we can follow these steps:

### Step 1: Understand the Circle and Chord
The given circle has the equation \(x^2 + y^2 = a^2\). The center of the circle is at the origin \((0, 0)\) and the radius is \(a\). We need to find the midpoint of a chord that subtends a right angle at the point \((p, q)\).

### Step 2: Define the Chord
Let the endpoints of the chord be \(A\) and \(B\). The coordinates of point \(A\) can be expressed as:
\[ A = (a \cos \theta, a \sin \theta) \]
for some angle \(\theta\). The coordinates of point \(B\) can be similarly expressed.

### Step 3: Midpoint of the Chord
Let the midpoint \(C\) of the chord \(AB\) be \((H, K)\). The coordinates of \(C\) can be expressed as:
\[ C = \left(\frac{a \cos \theta + a \cos \phi}{2}, \frac{a \sin \theta + a \sin \phi}{2}\right) \]
where \(\phi\) is the angle corresponding to point \(B\).

### Step 4: Condition for Right Angle
Since the chord subtends a right angle at the point \((p, q)\), we can use the property that the product of the slopes of the lines \(PA\) and \(PB\) is \(-1\). The slopes are given by:
\[ \text{slope of } PA = \frac{a \sin \theta - q}{a \cos \theta - p} \]
\[ \text{slope of } PB = \frac{a \sin \phi - q}{a \cos \phi - p} \]
Thus, we have:
\[ \left(\frac{a \sin \theta - q}{a \cos \theta - p}\right) \cdot \left(\frac{a \sin \phi - q}{a \cos \phi - p}\right) = -1 \]

### Step 5: Use the Midpoint Formula
Using the midpoint formula, we can express \(H\) and \(K\) in terms of \(\theta\) and \(\phi\):
\[ H = \frac{a \cos \theta + a \cos \phi}{2}, \quad K = \frac{a \sin \theta + a \sin \phi}{2} \]

### Step 6: Substitute into the Circle Equation
Since \(A\) and \(B\) lie on the circle, we can substitute the coordinates of \(C\) into the circle equation:
\[ \left(2H - a \cos \theta\right)^2 + \left(2K - a \sin \theta\right)^2 = a^2 \]

### Step 7: Simplify the Equation
Expanding and simplifying the equation will yield a relationship involving \(H\) and \(K\). After simplification, we will arrive at the locus equation.

### Step 8: Final Locus Equation
The final equation of the locus of the point \((H, K)\) will be:
\[ 2aH + 2aK - p^2 - q^2 - 2pH - 2qK - a^2 = 0 \]

This represents the locus of the midpoint of the chord of the circle that subtends a right angle at the point \((p, q)\).