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

Answer: (C)

6

Answer 2

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:

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

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