Animated Solution for Mathematics - Circles: The centre of the circle passing through $(0,0)$ and $(1,0)$ and touching the circle $x^2+y^2=9$ is

A Journey to the Center

Welcome, future engineer! Today, we are not just solving a coordinate geometry problem; we are peeling back the layers of a geometric puzzle. When you look at a problem like this, it is easy to get lost in the algebra.

But I want you to pause. Take a deep breath. Geometry is not about equations; it is about visualization. Let us walk through this together.

The Setup

We are given a fixed circle, $x^2+y^2=9$. Immediately, your brain should register the vital statistics: the center is at the origin $C_2(0,0)$ and the radius $r_2$ is $3$.

Now, we are tasked with finding a new circle that passes through the origin $(0,0)$ and the point $(1,0)$. Think about this. Our new circle is 'anchored' at the origin. It is pinned to the very heart of the fixed circle. This is the first clue that this is not a standard intersection problem; it is a tangency problem.

The General Equation

To find the equation of our mystery circle, we start with the most powerful tool in our arsenal: the general equation of a circle:

Our goal is to find the constants $g$, $f$, and $c$. We have two points, $(0,0)$ and $(1,0)$, which act as our keys to unlock these constants.

First, we substitute $(0,0)$ into the equation. As expected, the terms vanish, leaving us with $c=0$. One down, two to go!

Next, we use the point $(1,0)$. Substituting $x=1$ and $y=0$ gives us:

This simplifies beautifully to $1+2g=0$, which means $g=-\frac{1}{2}$. We are making incredible progress.

We now know that our center is $(-g,-f)=(\frac{1}{2},-f)$ and our radius is $r_1=\sqrt{g^2+f^2-c}=\sqrt{\frac{1}{4}+f^2}$.

The Tangency Trap

Here is where most students stumble. We know the circles touch. But how?

Because our circle passes through the center of the fixed circle, it must lie entirely within it. This is the definition of internal tangency. For two circles to touch internally, the distance between their centers, $C_1{C}_2$, must be equal to the absolute difference of their radii:

Let us calculate the distance between the centers. $C_1$ is $(\frac{1}{2},-f)$ and $C_2$ is $(0,0)$. The distance is:

Wait! Look closely. This distance is exactly equal to our radius $r_1$. So, our condition becomes:

The Final Calculation

Since $r_1$ must be positive and less than $3$ (because it is inside the larger circle), we can drop the absolute value sign:

Now, we equate this to our expression for the radius:

Squaring both sides gives us $\frac{1}{4}+f^2=\frac{9}{4}$. Subtracting $\frac{1}{4}$ from both sides, we get $f^2=\frac{8}{4}=2$. Therefore, $f=\pm \sqrt{2}$.

Our center is $(-g,-f)=(\frac{1}{2},-f)$. Substituting our values, the center is $(\frac{1}{2},-\sqrt{2})$ or $(\frac{1}{2},\sqrt{2})$.

Looking at our options, we find the perfect match: $(\frac{1}{2},-\sqrt{2})$.

See? When you break it down, the complexity melts away. You have mastered the geometry, navigated the tangency condition, and solved the algebra. Keep this confidence, and you will conquer any problem the JEE throws at you!