The circle passing through the intersection of the circles, x² + y² - 6x = 0 and x² + y² - 4y = 0 ha

Q: Why do we use the family of circles equation $S_1 + \lambda S_2 = 0$? Can't we just find the intersection points?

You absolutely could, but in JEE Advanced, time is your most precious currency. Finding the intersection points involves solving a system of quadratic equations, which is prone to calculation errors and takes valuable time. The family of circles method is elegant, direct, and keeps the geometry intact without getting lost in the weeds of coordinate calculation. It is the 'pro' way to handle intersection problems.

Q: Wait, what happens if $1 + \lambda = 0$? Is that a valid circle?

That is a brilliant observation! If $1 + \lambda = 0$, the $x^2$ and $y^2$ terms vanish, and the equation becomes linear. This represents the 'radical axis'—the common chord passing through the intersection points. It is a degenerate circle with an infinite radius, essentially a straight line. Since we are looking for a circle, we assume $1 + \lambda \neq 0$.

Q: Why do we divide by $(1 + \lambda)$ before finding the center?

The standard form of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ requires the coefficients of $x^2$ and $y^2$ to be unity (one). If you don't normalize the equation, your center coordinates will be wrong, and the entire calculation will collapse. Always normalize first to ensure your $g$ and $f$ values are correct!

Question

The circle passing through the intersection of the circles, $x^2 + y^2 - 6x = 0$ and $x^2 + y^2 - 4y = 0$ having its centre on the line, $2x - 3y + 12 = 0$, also passes through the point:

Options:

(A) $(-3, 6)$

(B) $(-1, 3)$

(C) $(-3, 1)$

(D) $(1, -3)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you have two circles, $S_1: x^2 + y^2 - 6x = 0$ and $S_2: x^2 + y^2 - 4y = 0$. They intersect at two points.

Instead of solving for these points directly, we utilize the **Family of Circles** theorem. This theorem states that any circle passing through the intersection of $S_1$ and $S_2$ can be represented by the equation $S_1 + \lambda S_2 = 0$, where $\lambda$ is a parameter.

Think of $\lambda$ as a dial. As you turn this dial, you sweep through an infinite collection of circles, all sharing the same two intersection points. Our task is to find the specific value of $\lambda$ that places the center of our circle on the line $2x - 3y + 12 = 0$.

### The Algebraic Grind

Let us write out our equation:
$$(x^2 + y^2 - 6x) + \lambda (x^2 + y^2 - 4y) = 0$$

Now, we organize the terms:
$$(1 + \lambda)x^2 + (1 + \lambda)y^2 - 6x - 4\lambda y = 0$$

To find the center, we must normalize the coefficients of $x^2$ and $y^2$ to $1$. Dividing the entire equation by $(1 + \lambda)$, we obtain:
$$x^2 + y^2 - \frac{6}{1+\lambda}x - \frac{4\lambda}{1+\lambda}y = 0$$

Comparing this to the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$, we identify the center coordinates as $(-g, -f)$. This yields the general center $C$:
$$C = \left( \frac{3}{1+\lambda}, \frac{2\lambda}{1+\lambda} \right)$$

### The Constraint: The Bridge to the Solution

We are given the constraint that the center must lie on the line $2x - 3y + 12 = 0$. Since the center lies on this line, its coordinates must satisfy the line's equation.

Substituting our $x$ and $y$ values into the line equation:
$$2\left( \frac{3}{1+\lambda} \right) - 3\left( \frac{2\lambda}{1+\lambda} \right) + 12 = 0$$

Multiplying by $(1 + \lambda)$ to clear the denominator, we get:
$$6 - 6\lambda + 12(1 + \lambda) = 0$$

Expanding this expression:
$$6 - 6\lambda + 12 + 12\lambda = 0 \Rightarrow 6\lambda + 18 = 0$$

Solving for the parameter, we find **$\lambda = -3$**.

### The Victory

Now, we substitute $\lambda = -3$ back into our grouped equation:
$$(1 - 3)x^2 + (1 - 3)y^2 - 6x - 4(-3)y = 0$$

This simplifies to:
$$-2x^2 - 2y^2 - 6x + 12y = 0$$

Dividing by $-2$, we arrive at the final equation of our circle:
**$$x^2 + y^2 + 3x - 6y = 0$$**

To verify, we test the point $(-3, 6)$:
$$(-3)^2 + (6)^2 + 3(-3) - 6(6) = 9 + 36 - 9 - 36 = 0$$

The point satisfies the equation perfectly. You have successfully navigated the problem by leveraging the geometric significance of the family of circles.

Animated Solution for Mathematics - Circles: The circle passing through the intersection of the circles, $x^{2} + y^{2} - 6 x = 0$ and $x^{2} + y^{2} - 4 y = 0$ having its centre on the line, $2 x - 3 y + 12 = 0$ , also passes through the point:

Select Answer:

Visualized Solution

Identify the Given Circles

The Family of Circles Concept

Substitute the Equations

Group the Terms

Convert to Standard Form

Identify the Center Coordinates

Apply the Line Constraint

Solve for $λ$

Find the Final Circle Equation

Verify the Given Options

The Sigma Insight: Director Circle and Family of Circles

Analyzing the Setup

The Algebraic Grind

The Constraint

The Victory

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