Two chords of a circle, of lengths 2a and 2b are mutually perpendicular. If the distance of the point, at which the chords interect, from the centre of the circle is c (c `lt` radius of the circle), then the radius of the circle is :

Options

A. `sqrt((a^(2) + b^(2) + c^(2))/(2))`
B. `(sqrt(ab))/(c)`
C. a + b - c
D. `(sqrt(a^(2) + b^(2) - c^(2)))/(2)`

Answer

To solve the problem, we need to find the radius of the circle given that two chords of lengths \(2a\) and \(2b\) are mutually perpendicular, and the distance from the center of the circle to the point of intersection of the chords is \(c\).

### Step-by-Step Solution:

1. **Understand the Geometry**:
- Let \(O\) be the center of the circle.
- Let \(P\) be the point where the two chords intersect.
- The lengths of the chords are \(AB = 2a\) and \(CD = 2b\).
- Since the chords are perpendicular, we can consider the right triangles formed.

2. **Set Up the Right Triangles**:
- The distance from the center \(O\) to the point \(P\) is given as \(c\).
- The distance from \(P\) to the endpoints of the chord \(AB\) is \(AP = PB = a\).
- The distance from \(P\) to the endpoints of the chord \(CD\) is \(CP = PD = b\).

3. **Use the Pythagorean Theorem**:
- For chord \(AB\):
\[ OA^2 = OP^2 + AP^2 \]
\[ OA^2 = c^2 + a^2 \]
- For chord \(CD\):
\[ OC^2 = OP^2 + CP^2 \]
\[ OC^2 = c^2 + b^2 \]

4. **Equate the Radii**:
- Since \(OA\) and \(OC\) are both radii of the circle, we can set them equal:
\[ c^2 + a^2 = c^2 + b^2 \]

5. **Simplify the Equation**:
- Cancel \(c^2\) from both sides:
\[ a^2 = b^2 \]
- This implies \(a = b\) or \(a = -b\) (which is not relevant in this context since lengths are positive).

6. **Find the Radius**:
- Now, we can express the radius \(r\) in terms of \(a\) and \(c\):
\[ r^2 = c^2 + a^2 \]
- Thus, the radius \(r\) of the circle is:
\[ r = \sqrt{c^2 + a^2} \]

### Final Answer:
The radius of the circle is:
\[ r = \sqrt{c^2 + a^2} \]