Find the locus of the middle points of chords of the circle `x^(2) + y^(2) = a^(2)` Which subtend a right angle at the point (c, 0)

Answer

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

### Step 1: Define the Circle and Points
The given circle is defined by the equation:
\[ x^2 + y^2 = a^2 \]
Let \(P(h, k)\) be the midpoint of a chord \(AB\) of the circle that subtends a right angle at the point \(C(c, 0)\).

### Step 2: Use the Midpoint Formula
The coordinates of points \(A\) and \(B\) on the circle can be represented parametrically as:
\[ A = (a \cos \theta, a \sin \theta) \quad \text{and} \quad B = (a \cos \phi, a \sin \phi) \]
The midpoint \(P(h, k)\) of the chord \(AB\) is given by:
\[ h = \frac{a \cos \theta + a \cos \phi}{2}, \quad k = \frac{a \sin \theta + a \sin \phi}{2} \]

### Step 3: Condition for Right Angle
For the chord \(AB\) to subtend a right angle at point \(C(c, 0)\), the slopes of lines \(CA\) and \(CB\) must satisfy the condition:
\[ m_{CA} \cdot m_{CB} = -1 \]
Calculating the slopes:
\[ m_{CA} = \frac{a \sin \theta - 0}{a \cos \theta - c} = \frac{a \sin \theta}{a \cos \theta - c} \]
\[ m_{CB} = \frac{a \sin \phi - 0}{a \cos \phi - c} = \frac{a \sin \phi}{a \cos \phi - c} \]
Thus, we have:
\[ \frac{a \sin \theta}{a \cos \theta - c} \cdot \frac{a \sin \phi}{a \cos \phi - c} = -1 \]

### Step 4: Substitute and Simplify
Cross-multiplying gives:
\[ a^2 \sin \theta \sin \phi = -(a \cos \theta - c)(a \cos \phi - c) \]
Expanding the right-hand side:
\[ a^2 \sin \theta \sin \phi = -a^2 \cos \theta \cos \phi + ac(\cos \theta + \cos \phi) - c^2 \]

### Step 5: Use Trigonometric Identities
Using the identities \(\sin \theta \sin \phi = \frac{1}{2}(\cos(\theta - \phi) - \cos(\theta + \phi))\) and \(\cos \theta \cos \phi = \frac{1}{2}(\cos(\theta + \phi) + \cos(\theta - \phi))\), we can express the equation in terms of \(h\) and \(k\).

### Step 6: Rearranging
After substituting \(h\) and \(k\) back into the equation, we can rearrange to express it as:
\[ 2h^2 + 2k^2 + c^2 - a^2 - 2hc = 0 \]
This can be rearranged to:
\[ 2h^2 + 2k^2 - 2hc + (c^2 - a^2) = 0 \]

### Step 7: Final Form
To express this in a standard form, we can complete the square:
\[ 2(h^2 - hc) + 2k^2 = a^2 - c^2 \]
Dividing by 2 gives:
\[ (h - \frac{c}{2})^2 + k^2 = \frac{a^2 - c^2}{2} \]

### Conclusion
The locus of the midpoints of the chords of the circle that subtend a right angle at the point \((c, 0)\) is a circle centered at \((\frac{c}{2}, 0)\) with a radius of \(\sqrt{\frac{a^2 - c^2}{2}}\).