Animated Solution for Mathematics - Circles: The number of common tangents, to the circles $x^2+y^2-18x-15y+131=0$ and $x^2+y^2-6x-6y-7=0$, is

My dear student, welcome to the beautiful world of coordinate geometry. Today, we are not just solving a problem; we are exploring the intimate relationship between two circles.

When you look at the equations $x^2+y^2-18x-15y+131=0$ and $x^2+y^2-6x-6y-7=0$, do not see them as mere collections of numbers. See them as two distinct entities, two circular orbits, waiting to be understood.

Our goal is to find the number of common tangents, but to do that, we must first understand how these circles relate to each other in space.

Every circle has a story, and that story is told by its center and its radius. We are given the equations in the general form $x^2+y^2+2gx+2fy+c=0$.

To find the center, we look at the coefficients of $x$ and $y$. For the first circle, $C_1$, the equation is $x^2+y^2-18x-15y+131=0$. By comparing this to the general form, we find the center $C_1=(9,\frac{15}{2})$.

Now, for the radius $r_1$, we use the elegant formula $r=\sqrt{g^2+f^2-c}$. Substituting our values, we get:

We repeat this process for the second circle, $C_2$, with the equation $x^2+y^2-6x-6y-7=0$. The center is $C_2=(3,3)$.

The radius $r_2$ is calculated as:

Now we have the vital statistics of our two circles: $C_1(9,7.5)$ with $r_1=2.5$, and $C_2(3,3)$ with $r_2=5$.

Now, we must bridge the gap between these two centers. The distance $d$ between $C_1(9,7.5)$ and $C_2(3,3)$ is given by the distance formula:

This simplifies to:

Or, in fractional form, $d=\frac{15}{2}$.

Here is where the magic happens. We compare the distance between the centers, $d=\frac{15}{2}$, with the sum of the radii, $r_1+r_2=\frac{5}{2}+5=\frac{15}{2}$.

Look at that! $d=r_1+r_2$. This is not a coincidence; it is a geometric revelation.

When the distance between the centers is exactly equal to the sum of the radii, the circles are perfectly touching each other externally. They are, in a sense, kissing at a single point.

Because they touch externally, we can visualize the tangents. We have two direct common tangents that run along the outer edges of both circles.

And, because they touch at a single point, we have one transverse common tangent that passes right through that point of contact.

That gives us a total of $2+1=3$ common tangents. It is a beautiful, clean result, born from the simple, elegant logic of geometry. Keep this visualization in your mind, and you will never fear circle problems again. The final answer is 3.