Animated Solution for Mathematics - Circles: The circle passing through $(1,-2)$ and touching the axis of $x$ at $(3,0)$ also passes through the point

Imagine you are standing on a vast, flat coordinate plane. You have a circle, a perfect, elegant curve, and it is performing a delicate dance with the x-axis. It doesn't cross the axis; it just kisses it, touching it at exactly one point: $(3,0)$.

In coordinate geometry, this isn't just a visual detail; it is a powerful constraint that unlocks the entire problem. Because the x-axis is a horizontal line, the normal to the circle at $(3,0)$ must be a vertical line.

This means the center of our circle, let's call it $C$, must lie somewhere on the vertical line $x=3$. We have just reduced our search space from the entire plane to a single line.

Since our center $C$ lies on the line $x=3$, we can define its coordinates as $(3,r)$, where $r$ is the unknown y-coordinate. Now, think about the radius $R$.

The distance from the center $(3,r)$ to the point of tangency $(3,0)$ is the radius of the circle. Using the distance formula, this is simply:

So, the radius $R$ is $|r|$. This is the beauty of tangency: it links the center's position directly to the radius. We don't need two separate variables for the center and radius; we only need $r$.

With the center at $(3,r)$ and the radius $R=|r|$, we can write the standard equation of the circle:

Substituting $R^2=r^2$, we get the elegant equation:

This equation describes every circle that touches the x-axis at $(3,0)$. Now, we have one final piece of information: the circle passes through the point $(1,-2)$.

This point must satisfy our equation. Let's substitute $x=1$ and $y=-2$ into the equation:

Now, let's expand this. $(1-3{)}^2$ is $(-2{)}^2$, which is $4$. The term $(-2-r{)}^2$ is the same as $(2+r{)}^2$, which expands to $4+4r+r^2$.

Putting it all together, we have:

Look at that! The $r^2$ terms on both sides cancel out, leaving us with a simple linear equation:

Solving for $r$, we get $4r=-8$, so $r=-2$. Our center is $(3,-2)$ and our radius is $2$. The equation of our circle is:

We have our circle. Now, we test the options to see which point lies on it. We test $(5,-2)$:

Since $4=4$, the point $(5,-2)$ lies on the circle. We have solved it!

This problem teaches us that geometry is not just about shapes; it is about constraints. Every piece of information given—the tangency, the points—is a tool to reduce the complexity of the problem. Keep this in mind, and you will find that even the most daunting problems have a path to the final answer.