Animated Solution for Mathematics - Circles: If $P$ and $Q$ are the points of intersection of the circles $x^2+y^2+3x+7y+2p-5=0$ and $x^2+y^2+2x+2y-p^2=0$ then there is a circle passing through $P,Q$ and $(1,1)$ for:

A Tale of Two Circles

Imagine you are standing on a vast coordinate plane. Before you lie two circles, $S_1$ and $S_2$. They are not just sitting there; they are locked in an embrace, intersecting at two distinct points, $P$ and $Q$.

You are tasked with finding a third circle—a circle that passes through these two intersection points and a specific coordinate, $(1,1)$. This sounds like a daunting task, but let's peel back the layers of this problem together.

The Common Chord

When two circles intersect, they share a secret: the line passing through their intersection points, $P$ and $Q$. We call this the common chord, $L$.

We do not need to solve for $P$ and $Q$ individually. Instead, we use the power of subtraction. By subtracting the equation of one circle from the other, $S_1-S_2=0$, the quadratic terms $x^2$ and $y^2$ vanish, leaving us with a linear equation.

Given our circles:

Subtracting $S_2$ from $S_1$ gives us the common chord $L$:

This line $L$ is the backbone of our construction.

The Family of Circles

Now, here is the magic. Any circle passing through the intersection of $S_1$ and $S_2$ can be represented by the family of circles equation:

Here, $\lambda$ is our parameter. By varying $\lambda$, we can generate an infinite number of circles that all pass through $P$ and $Q$. We are looking for the one specific circle that also passes through the point $(1,1)$.

The Constraint of the Point

Since our desired circle must pass through $(1,1)$, we simply substitute $x=1$ and $y=1$ into our family equation. Let's evaluate $S_1$ and $L$ at this point.

Evaluating $S_1(1,1)$:

Evaluating $L(1,1)$:

Substituting these back into $S_1+\lambda L=0$:

Solving for $\lambda$:

Look at that denominator: $(p+1{)}^2$. In the realm of mathematics, division by zero is the ultimate forbidden act. If $p=-1$, the denominator vanishes, and $\lambda$ becomes undefined.

This isn't just a calculation error; it's a geometric reality. When $p=-1$, the point $(1,1)$ lies directly on the common chord $L$. You cannot draw a circle through two points and a third point if all three are collinear!

Thus, for every value of $p$ except $p=-1$, we can find a unique $\lambda$, and therefore, a unique circle. We have conquered the problem. The answer is: all real values of $p$ except $p=-1$.