Question

If one of the diameters of the circle \( x^2 + y^2 - 10x + 4y + 13 = 0 \) is a chord of another circle \( C \), whose center is the point of intersection of the lines \( 2x + 3y = 12 \) and \( 3x - 2y = 5 \),then the radius of the circle \( C \) is

• A. \( \sqrt{20} \)
• B. 4
• C. 6
• D. \( 3 \sqrt{2} \)

Answer 1

The Correct Option is C

To find the center of circle \( C \), consider the intersection of the lines:

\[ 2x + 3y = 12 \quad \text{and} \quad 3x - 2y = 5 \]

Solving these equations:

\[ 13x = 39 \quad \implies \quad x = 3, \; y = 2 \]

Therefore, the center of the circle is at:

\[ (3, 2) \]

Given circle equation:

\[ x^2 + y^2 - 10x + 4y + 13 = 0 \]

The center of this circle is at \( (5, -2) \) and its radius is:

\[ \sqrt{(5)^2 + (-2)^2 - 13} = \sqrt{25 + 4 - 13} = 4 \]

Calculate distances:

\[ CM = \sqrt{(3 - 5)^2 + (2 - (-2))^2} = \sqrt{4 + 16} = 5\sqrt{2} \]

\[ CP = \sqrt{(3 - 5)^2 + (2 - 0)^2} = \sqrt{16 + 20} = 6 \]