Question

A circle passes through the point \( (6,2) \). If segments of the straight lines \( x+y=6 \) and \( x+2y=4 \) are two diameters of the circle, then its radius is

• A. \( 4 \)
• B. \( 8 \)
• C. \( \sqrt{5} \)
• D. \( 2\sqrt{5} \)
• E. \( 4\sqrt{5} \)

Hint:

Whenever diameters are given as lines, their intersection always gives the center of the circle.

Answer 1

The Correct Option is D

Concept:
• The center of a circle lies at the midpoint of any diameter.
• If two diameters are given as line segments, their intersection point gives the center of the circle.
• Radius is the distance from center to any point on the circle.

Step 1: Finding the center of the circle.
The diameters lie along the lines: \[ x + y = 6 \quad ...(1) \] \[ x + 2y = 4 \quad ...(2) \] The center is the intersection point of these lines.

Step 2: Solving the equations.
Subtract (1) from (2): \[ (x + 2y) - (x + y) = 4 - 6 \] \[ y = -2 \] Substitute into (1): \[ x - 2 = 6 \Rightarrow x = 8 \] Thus, center: \[ C = (8,-2) \]

Step 3: Using given point to find radius.
Point on circle: \[ P = (6,2) \] Radius = distance \( CP \): \[ r = \sqrt{(6-8)^2 + (2+2)^2} \]

Step 4: Simplifying.
\[ r = \sqrt{(-2)^2 + (4)^2} \] \[ r = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]

Step 5: Final Answer.
\[ \boxed{2\sqrt{5}} \]