Consider a family of circles passing through two fixed points A(3, 7) and B(6, 5). Show that the cho

Q: Wait, why do we use $S + \lambda L = 0$ instead of $S_1 + \lambda S_2 = 0$?

Great question! While $S_1 + \lambda S_2 = 0$ is mathematically valid, it is often computationally heavier. Using the line $L$ (the radical axis of the family) simplifies the algebra significantly because $L$ is linear. It makes the subsequent subtraction much cleaner and keeps the degree of the equation manageable.

Q: Why does subtracting the equations of two circles give the common chord?

Think of it as the 'Radical Axis'. The radical axis is the locus of points where the power of the point with respect to both circles is equal. Algebraically, when you subtract $S_1 - S_2 = 0$, the quadratic terms $x^2$ and $y^2$ vanish, leaving you with a linear equation—the equation of a line. That line is the common chord!

Q: Is the point of concurrency always inside the fixed circle?

Not necessarily! The point of concurrency is a property of the family of circles and the fixed circle. It is the intersection of two specific lines derived from the geometry. It can lie anywhere on the plane depending on the relative positions of the fixed points and the fixed circle.

Question

Consider a family of circles passing through two fixed points $A(3, 7)$ and $B(6, 5)$. Show that the chords in which the circle $x^2 + y^2 - 4x - 6y - 3 = 0$ cuts the members of the family are concurrent at a point. Find the coordinate of this point.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two fixed points, $A(3, 7)$ and $B(6, 5)$. These points are the anchors of our universe today.

We are not just looking at one circle; we are looking at an infinite family of circles that pass through these two points. It is a beautiful, breathing structure.

Our mission is to intersect this family with a fixed circle, $C_0: x^2 + y^2 - 4x - 6y - 3 = 0$, and prove that the resulting common chords are not chaotic—they are perfectly concurrent.

### The Power of the Family

To master this, we must use the most elegant tool in our arsenal: the family of circles equation, $S + \lambda L = 0$. Here, $L=0$ is the line passing through $A$ and $B$.

We calculate the slope of $AB$ as:

$$m = \frac{5-7}{6-3} = -\frac{2}{3}$$

Using the point-slope form, we derive the line $L$:

$$2x + 3y - 27 = 0$$

For our base circle $S$, we choose the one where $AB$ is the diameter, giving us:

$$(x-3)(x-6) + (y-7)(y-5) = 0$$

This simplifies to:

$$x^2 + y^2 - 9x - 12y + 53 = 0$$

Now, our family is defined by:

$$S_\lambda: (x^2 + y^2 - 9x - 12y + 53) + \lambda(2x + 3y - 27) = 0$$

This equation represents every possible circle passing through $A$ and $B$.

### The Radical Axis Revelation

Now, we introduce the fixed circle $C_0$. When we intersect any circle from our family $S_\lambda$ with $C_0$, the common chord is found by simply subtracting their equations: $S_\lambda - C_0 = 0$.

Watch the magic happen: the quadratic terms $x^2$ and $y^2$ cancel out perfectly! We are left with:

$$[(x^2 + y^2 - 9x - 12y + 53) + \lambda(2x + 3y - 27)] - (x^2 + y^2 - 4x - 6y - 3) = 0$$

Grouping the terms, we get:

$$(-5x - 6y + 56) + \lambda(2x + 3y - 27) = 0$$

### The Final Concurrency

Look at the structure of this equation. It is in the form $L_1 + \lambda L_2 = 0$, where $L_1: 5x + 6y - 56 = 0$ and $L_2: 2x + 3y - 27 = 0$.

This is the classic equation of a family of lines passing through the intersection of $L_1$ and $L_2$. No matter what value $\lambda$ takes, the common chord must pass through this intersection point.

Solving the system $5x + 6y = 56$ and $2x + 3y = 27$ (or $4x + 6y = 54$), we subtract to find $x = 2$. Substituting back, we find $y = \frac{23}{3}$.

We have proven that all these chords meet at the point **$(2, \frac{23}{3})$**. Geometry is not just about shapes; it is about the hidden order beneath the surface. You have just uncovered it.

Visualized Solution

The Geometric Setup

Family of Circles Concept

Equation of Line $L$

Equation of Circle $S$

The Family Equation $S_{λ}$

The Common Chord Equation

Subtracting the Circles

Grouping the Terms

Family of Lines Concept

Solving for the Concurrency Point

Final Coordinates

The Sigma Insight: Director Circle and Family of Circles

Analyzing the Setup

The Power of the Family

The Radical Axis Revelation

The Final Concurrency

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