Two circles x² + y² = 6 and x² + y² - 6x + 8 = 0 are given. Then the equation of the circle through

Q: Why can't we just find the intersection points of the two circles and then find the equation of the circle passing through three points?

You absolutely could! But in JEE Advanced, time is your most precious currency. Solving for intersection points involves solving a system of quadratic equations, which is prone to calculation errors. The 'Family of Circles' method is elegant, faster, and much less likely to lead to a sign error. It is the 'pro' way to handle this problem.

Q: Why does the equation $S_1 + \lambda(S_1 - S_2) = 0$ represent a circle?

Great question! If you expand this, you get $(1+\lambda)x^2 + (1+\lambda)y^2 + \dots = 0$. As long as $\lambda \neq -1$, the coefficients of $x^2$ and $y^2$ are equal and non-zero, which is the hallmark of a circle. It is a beautiful algebraic structure that preserves the circular nature of the equation!

Q: What if the radical axis is not a line?

The radical axis is *always* a line for two circles. Because the $x^2$ and $y^2$ terms in the equations of two circles always have the same coefficients (usually 1), subtracting one from the other will *always* eliminate the quadratic terms, leaving you with a linear equation. It is a geometric guarantee!

Question

Two circles $x^2 + y^2 = 6$ and $x^2 + y^2 - 6x + 8 = 0$ are given. Then the equation of the circle through their points of intersection and the point $(1, 1)$ is

Options:

(A) $x^2 + y^2 - 6x + 4 = 0$

(B) $x^2 + y^2 - 3x + 1 = 0$

(C) $x^2 + y^2 - 4y + 2 = 0$

(D) none of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two circles. One is centered at the origin, calm and symmetric. The other is shifted, centered at $(3,0)$.

They intersect, creating two distinct points of contact. You are asked to find the equation of a new circle that passes through these two points, and also through the point $(1,1)$.

This is not just an algebra problem; it is a problem of understanding the 'DNA' of circles. In the world of coordinate geometry, we have a powerful tool called the **Family of Circles**. Think of this as a way to describe an infinite number of circles that all share a common 'spine'—the line passing through their intersection points.

### The Power of the Family

When we write the equation of a circle as $S_1 = 0$ and another as $S_2 = 0$, we are defining their boundaries. If we want a circle that passes through their intersection, we don't need to find the points themselves.

Instead, we use the linear combination:
$$S_1 + \lambda(S_1 - S_2) = 0$$

Why does this work? If a point lies on both $S_1$ and $S_2$, then $S_1 = 0$ and $S_2 = 0$. Consequently, the entire expression becomes $0 + \lambda(0 - 0) = 0$.

It is a mathematical identity that holds true for any value of $\lambda$. This $\lambda$ is our 'tuning knob.' By changing $\lambda$, we can slide through the infinite family of circles that share the same intersection points.

### Unveiling the Radical Axis

The term $(S_1 - S_2)$ is the secret key. When we subtract the equation of one circle from another, the quadratic terms $x^2$ and $y^2$ vanish. What remains is a linear equation: $L = 0$.

This is the **Radical Axis**, the common chord of the two circles. Let us perform this calculation. We have:
$$S_1: x^2 + y^2 - 6 = 0$$
$$S_2: x^2 + y^2 - 6x + 8 = 0$$

Subtracting $S_2$ from $S_1$ gives us:
$$(x^2 + y^2 - 6) - (x^2 + y^2 - 6x + 8) = 0$$

The $x^2$ and $y^2$ terms cancel out beautifully, leaving us with:
$$6x - 14 = 0$$

This is our Radical Axis. It is the straight line that acts as the backbone for our family of circles.

### The Final Assembly

Now, we construct our family equation:
$$(x^2 + y^2 - 6) + \lambda(6x - 14) = 0$$

We are almost there. We have an infinite family, but we need the one specific circle that passes through $(1,1)$. We substitute $x=1$ and $y=1$ into our equation:
$$(1^2 + 1^2 - 6) + \lambda(6(1) - 14) = 0$$

Simplifying this, we get $(2 - 6) + \lambda(6 - 14) = 0$, which leads to $-4 - 8\lambda = 0$. Solving for $\lambda$, we find:
$$\lambda = -\frac{1}{2}$$

This is the specific value that selects our unique circle from the infinite family.

### The Result

Finally, we substitute $\lambda = -\frac{1}{2}$ back into our family equation:
$$(x^2 + y^2 - 6) - \frac{1}{2}(6x - 14) = 0$$

Expanding this, we get:
$$x^2 + y^2 - 6 - 3x + 7 = 0$$

Combining the constants, we arrive at the final, elegant equation:
$$x^2 + y^2 - 3x + 1 = 0$$

This is the circle you were looking for. It passes through the intersection points and the point $(1,1)$.

Notice how the process flowed: we identified the family, found the radical axis, applied the constraint, and solved for the parameter. This is the essence of **JEE Advanced** problem-solving—not just grinding through calculations, but using the elegant structure of geometry to find the path of least resistance.

Animated Solution for Mathematics - Circles: Two circles $x^{2} + y^{2} = 6$ and $x^{2} + y^{2} - 6 x + 8 = 0$ are given. Then the equation of the circle through their points of intersection and the point $(1, 1)$ is

Select Answer:

Visualized Solution

Visualizing the Given Circles

The Family of Circles Concept

Understanding the Radical Axis

Calculating the Radical Axis $L$

Simplifying the Line Equation

Setting up the Family Equation

Introducing the Point $(1, 1)$

Substituting the Point $(1, 1)$

Solving for $λ$

Substituting $λ$ Back

The Final Equation

Conclusion and Verification

The Sigma Insight: Director Circle and Family of Circles

Analyzing the Setup

The Power of the Family

Unveiling the Radical Axis

The Final Assembly

The Result

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