Animated Solution for Mathematics - Circles: The equation of the circle passing through $(1,1)$ and the points of intersection of $x^2+y^2+13x-3y=0$ and $2x^2+2y^2+4x-7y-25=0$ is

When you look at two circles, $S_1:x^2+y^2+13x-3y=0$ and $S_2:2x^2+2y^2+4x-7y-25=0$, the expert sees a 'Family of Circles'. These two circles intersect at two points, $A$ and $B$, and there are infinitely many circles that pass through these points.

Instead of solving for the coordinates of $A$ and $B$—which is algebraically tedious—we use the Family of Circles theorem. This theorem states that any circle passing through the intersection of $S_1=0$ and $S_2=0$ can be expressed as:

Here, $\lambda$ is a parameter that allows us to sweep through every possible circle passing through the intersection points.

Our goal is to find the specific value of $\lambda$ that corresponds to the circle passing through the point $P(1,1)$. We set up our master equation as follows:

Since the point $P(1,1)$ lies on our target circle, its coordinates must satisfy this equation. We substitute $x=1$ and $y=1$ into the expression:

For $S_1$:

For $S_2$:

Substituting these values back into our master equation, we obtain:

This simplifies to the linear equation $12-24\lambda =0$. Solving for $\lambda$, we find:

Now, we substitute $\lambda =\frac{1}{2}$ back into our family equation:

To clear the fraction, we multiply the entire equation by $2$:

Expanding the terms, we get:

Combining like terms, we arrive at the final, elegant result:

By utilizing the structural properties of the geometry, we have successfully avoided the messy calculation of intersection points. This approach is the hallmark of a JEE Advanced strategy.