JEE Mathematics 2024 Question with Solution

Given: One diameter of the circle $x^2 + y^2 - 10x + 4y + 13 = 0$ is a chord of another circle $C$. The center of circle $C$ is the intersection point of $2x + 3y = 12$ and $3x - 2y = 5$.

Find: The radius of circle $C$.

First, write the given circle in standard form.

So the center of this circle is $M(5,-2)$ and its radius is $4$.

Now find the center of circle $C$ by solving

and

Multiply the first equation by $2$ and the second by $3$:

Adding,

Substitute in $2x + 3y = 12$:

Hence the center of circle $C$ is $C(3,2)$.

A diameter of the first circle is a chord of circle $C$. Any diameter passes through the center $M(5,-2)$, so the midpoint of that chord is $M$. Therefore, the perpendicular distance from the center $C(3,2)$ of circle $C$ to this chord is

The chord is a diameter of the first circle, so its half-length is equal to the radius of the first circle, namely $4$. For circle $C$,

Thus,

Therefore, the radius of circle $C$ is $6$. The correct option is C.

Given: The center of circle $C$ is the intersection of $2x + 3y = 12$ and $3x - 2y = 5$. A diameter of the circle $x^2 + y^2 - 10x + 4y + 13 = 0$ is a chord of circle $C$.

Find: Radius of circle $C$.

For the given circle,

so its center is $M(5,-2)$ and radius is $4$. Hence any diameter of this circle has length $8$.

The center of circle $C$ is obtained by solving the pair of lines, giving $C(3,2)$. The distance between $C(3,2)$ and $M(5,-2)$ is

Since the diameter of the first circle is a chord of circle $C$, its length in circle $C$ is also $8$. For a circle of radius $R$, if the perpendicular distance from the center to a chord is $d$ and half-chord length is $\ell$, then

Here $d = \sqrt{20}$ and $\ell = 4$, so

Hence the correct option is C.