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

Imagine you are standing on a vast, flat plane, looking down at a large, perfect circle defined by the equation $x^2+y^2=9$. This circle is centered at the origin, $O(0,0)$, and has a radius $R=3$.

We introduce a new, smaller circle into this scene. This smaller circle must pass through the origin $(0,0)$ and the point $(1,0)$, and it must touch the larger circle.

Let us start with the general equation of a circle: $x^2+y^2+2gx+2fy+c=0$. Because our circle passes through the origin $(0,0)$, we substitute these coordinates directly into the equation.

The result is immediate: $0^2+0^2+2g(0)+2f(0)+c=0$, which simplifies to $c=0$. We have successfully eliminated one variable.

Next, we know the circle passes through $(1,0)$. Substituting these values, we get $1^2+0^2+2g(1)+2f(0)=0$, which simplifies to $1+2g=0$.

Solving for $g$, we find $g=-\frac{1}{2}$. The $x$-coordinate of our center, which is $-g$, is now fixed at $\frac{1}{2}$.

Now, we address the condition of tangency. Our smaller circle passes through the center of the larger circle.

If you visualize this, you will realize that the smaller circle must be entirely contained within the larger one. This is a case of internal tangency.

The condition for two circles to touch internally is that the distance between their centers, $d$, must equal the difference of their radii: $d=|R-r|$. Here, $R=3$ and $r$ is the radius of our smaller circle.

Since the center of the smaller circle is $(-g,-f)=(\frac{1}{2},-f)$ and the center of the larger circle is $(0,0)$, the distance $d$ is simply the distance from the origin to the center of the smaller circle, which is $\sqrt{g^2+f^2}$.

However, since the circle passes through the origin, the distance from the center to the origin is exactly the radius $r$. Thus, the condition $d=R-r$ becomes $r=3-r$.

Solving this gives $2r=3$, or:

We know the radius $r$ is given by the formula $r=\sqrt{g^2+f^2-c}$. Since $r=\frac{3}{2}$, $g=-\frac{1}{2}$, and $c=0$, we can write:

This simplifies to:

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

The center of our circle is $(-g,-f)$. Substituting our values, we find the coordinates of the center are:

We have successfully navigated the geometry and the algebra to find the center of the required circle. The final coordinates for the center are $(\frac{1}{2},\sqrt{2})$ and $(\frac{1}{2},-\sqrt{2})$.