Animated Solution for Mathematics - Circles: Let $a$ and $b$ be non-zero real numbers. Then, the equation $(a{x}^2+b{y}^2+c)(x^2-5xy+6y^2)=0$ represents

Welcome, future engineers. Today, we are not just solving an equation; we are performing a dissection. In the realm of JEE Advanced, coordinate geometry is rarely about brute force. It is about pattern recognition, intuition, and the ability to see the hidden structure beneath a complex expression.

We are presented with the equation:

At first glance, it looks intimidating—a product of two quadratic expressions. But let us apply the most powerful tool in our algebraic arsenal: the Zero Product Property.

The Homogeneous Beast

When you see a product equal to zero, your mind should immediately split the problem into two distinct paths. The equation holds true if either the first factor is zero, or the second factor is zero. Let us focus on the second factor first:

Observe the degree of every term. The $x^2$ term has degree 2, the $xy$ term has degree $1+1=2$, and the $y^2$ term has degree 2. This is a homogeneous equation of degree 2, which represents a pair of straight lines passing through the origin $(0,0)$.

To see these lines, we factorize the expression. We look for two numbers that multiply to $6$ and add to $-5$, which are $-2$ and $-3$:

Grouping the terms, we get:

We have successfully deconstructed the beast into two lines: $L_1:x-2y=0$ and $L_2:x-3y=0$. Both pass through the origin with slopes of $\frac{1}{2}$ and $\frac{1}{3}$, respectively.

The Conic Section

Now, let us turn our attention to the first factor: $a{x}^2+b{y}^2+c=0$. This is the general form of a conic section centered at the origin.

What defines a circle in the Cartesian plane? It is the set of all points equidistant from a center. Algebraically, the coefficients of $x^2$ and $y^2$ must be equal. Thus, we impose the condition:

Substituting this into our equation, we get $a(x^2+y^2)+c=0$. Isolating the geometric core, we move $c$ to the right side and divide by $a$:

This is the standard form of a circle $x^2+y^2=r^2$, where the radius squared is $r^2=-\frac{c}{a}$.

The Radius Trap

Here is where the JEE examiners test your maturity. We know that for any real point $(x,y)$, the sum of squares $x^2+y^2$ must be greater than or equal to zero. Therefore, the right-hand side must satisfy:

A fraction is negative only when the numerator and denominator have opposite signs. This means $c$ and $a$ must have opposite signs. If $a$ is positive, $c$ must be negative; if $a$ is negative, $c$ must be positive.

The Synthesis

We have analyzed both parts of the original equation. The second factor yielded two straight lines intersecting at the origin. The first factor, under the conditions $a=b$ and $ac<0$, yields a circle.

When we combine these, the total locus is the union of these two shapes. You have successfully navigated the complexity, broken down the algebra, and verified the geometric constraints. This is the essence of coordinate geometry—not memorizing formulas, but understanding the behavior of equations.