Animated Solution for Mathematics - Circles: If one of the diameters of the circle $x^2+y^2-2x-6y+6=0$ is a chord to the circle with centre $(2,1)$, then the radius of the circle is

Welcome, fellow traveler in the realm of coordinate geometry! Today, we are going to dissect a problem that, at first glance, might seem like a standard algebraic exercise, but is actually a beautiful study in geometric relationships.

We are dealing with two circles, and the way they interact through a shared chord is a classic JEE Advanced concept. Let us peel back the layers.

We begin with the equation of our first circle, $C_1$:

To understand its soul, we must bring it into the standard form, $x^2+y^2+2gx+2fy+c=0$. By comparing coefficients, we find $2g=-2$, which gives us $g=-1$, and $2f=-6$, which gives us $f=-3$. The constant $c$ is $6$.

With these values, we can easily locate the center $C_1$ at $(-g,-f)$, which is $(1,3)$. The radius $r_1$ is calculated using the elegant formula:

Plugging in our numbers, we get:

So, we have a circle centered at $(1,3)$ with a radius of $2$. Keep this image in your mind; it is the anchor of our problem.

Now, the problem introduces a second circle, $C_2$, with its center at $(2,1)$. We are told that a diameter of $C_1$ acts as a chord for $C_2$. This is the "Aha!" moment.

Visualize this: a chord that is a diameter of $C_1$ must pass through the center of $C_1$. This chord is also a chord of $C_2$.

Recall the fundamental property of circles: the line segment connecting the center of a circle to the midpoint of a chord is always perpendicular to that chord. In our case, the center of $C_2$ is $(2,1)$, and the midpoint of the chord is the center of $C_1$, which is $(1,3)$.

Therefore, the line segment connecting $C_2(2,1)$ and $C_1(1,3)$ is perpendicular to the chord. This creates a right-angled triangle where the hypotenuse is the radius of the second circle, $R$.

We are almost there. We need the distance $d$ between the centers $C_1(1,3)$ and $C_2(2,1)$. Using the distance formula:

Substituting the coordinates:

Now, look at the right-angled triangle formed by the centers and the endpoint of the chord. The hypotenuse is $R$, the base is the radius of $C_1$ ($r_1=2$), and the perpendicular height is the distance $d=\sqrt{5}$.

By the Pythagorean theorem:

Substituting our values, we find:

Taking the square root, we get $R=3$.

And there it is! The radius of the second circle is $3$.

This problem teaches us that geometry is not just about equations; it is about seeing the hidden triangles and relationships between shapes. When you encounter such problems in the exam, do not rush to the algebra. Pause, visualize the centers, and look for the right-angled triangles.