A circle C touches the line x = 2y at the point (2, 1) and intersects the circle C1 : x² + y² + 2y -

Q: Wait, why do we use the family of circles equation $(x - x_1)^2 + (y - y_1)^2 + \lambda L = 0$ instead of just finding the center $(h, k)$ and radius $r$ directly?

That is a brilliant question! While you could define the center as $(h, k)$ and use the distance formula to the line and the point, it leads to a system of non-linear equations that are algebraically exhausting. The family of circles approach is the 'JEE shortcut'—it encapsulates the tangency condition into a single parameter $\lambda$, turning a geometric nightmare into a simple linear solve.

Q: Why does the common chord $PQ$ have to pass through the center of $C_1$?

Think about the definition of a diameter. A diameter is a chord that passes through the center of the circle. Since $PQ$ is a chord of $C_1$ and the problem explicitly states it is a diameter, it is geometrically mandatory that the center of $C_1$ lies on the line segment $PQ$.

Q: How do I know the equation $C - C_1 = 0$ represents the common chord?

When you subtract the equations of two circles, the $x^2$ and $y^2$ terms cancel out. What remains is a first-degree equation in $x$ and $y$, which is the equation of a line. Since this line satisfies both circle equations simultaneously, it must be the line passing through their intersection points $P$ and $Q$.

Q: I got $\lambda = -7$. Does the sign of $\lambda$ matter?

Absolutely! The sign of $\lambda$ determines the 'side' of the line on which the circle is formed and its curvature. In this specific family, $\lambda$ is a scaling factor that adjusts the circle's radius to satisfy the intersection condition. If you had gotten a positive value, it would simply mean the circle is on the other side of the tangent line.

Q: Is there any other way to find the diameter without calculating the full equation of $C$?

You could potentially use the power of a point or radical axis properties, but those are often more prone to calculation errors. The method we used—finding $\lambda$, then the center and radius—is the most robust and reliable path to the final answer.

Question

A circle $C$ touches the line $x = 2y$ at the point $(2, 1)$ and intersects the circle $C_1 : x^2 + y^2 + 2y - 5 = 0$ at two points $P$ and $Q$ such that $PQ$ is a diameter of $C_1$. Then the diameter of $C$ is :

Options:

(A) $7\sqrt{5}$

(B) $15$

(C) $\sqrt{285}$

(D) $4\sqrt{15}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane. You see a line, $x - 2y = 0$, and a specific point $(2, 1)$ resting on it. You are tasked with drawing a circle $C$ that kisses this line perfectly at that point.

This is not just any circle; it is a member of a vast family of circles, all sharing the same point of tangency. We use the elegant family of circles equation:

$$(x - 2)^2 + (y - 1)^2 + \lambda(x - 2y) = 0$$

By introducing the parameter $\lambda$, we have essentially created a 'knob' that we can turn to grow or shrink our circle until it perfectly satisfies the second condition of our problem.

### The Intersection: Where Two Worlds Meet

Now, we introduce a second circle, $C_1: x^2 + y^2 + 2y - 5 = 0$. This circle is already fixed in space with its center at $(0, -1)$. Our circle $C$ is destined to intersect $C_1$ at two points, $P$ and $Q$.

The problem gives us a beautiful geometric gift: the segment $PQ$ is a diameter of $C_1$. Because a diameter is the 'royal road' of the circle, it must pass directly through its center.

If $PQ$ is a diameter of $C_1$, then the line containing $PQ$ must pass through the center $(0, -1)$.

### The Power of Subtraction

To find the equation of the line $PQ$, we do not need to find the coordinates of $P$ and $Q$ individually. Instead, we use the radical axis theorem. By subtracting the equation of $C_1$ from our general equation of $C$, we eliminate the quadratic terms.

Expanding the general equation of $C$:
$$x^2 - 4x + 4 + y^2 - 2y + 1 + \lambda x - 2\lambda y = 0$$
$$x^2 + y^2 + (\lambda - 4)x - (2\lambda + 2)y + 5 = 0$$

Subtracting $C_1: x^2 + y^2 + 2y - 5 = 0$ from this result yields the equation of the common chord $PQ$:
$$(\lambda - 4)x - (2\lambda + 4)y + 10 = 0$$

Since we know this line must pass through the center of $C_1$, we substitute $(0, -1)$ into this equation:
$$(\lambda - 4)(0) - (2\lambda + 4)(-1) + 10 = 0$$
$$2\lambda + 4 + 10 = 0 \Rightarrow 2\lambda = -14 \Rightarrow \lambda = -7$$

### The Final Reveal

With $\lambda = -7$, our circle equation becomes:
$$x^2 + y^2 - 11x + 12y + 5 = 0$$

To find the diameter, we first identify the radius using the standard formula $R = \sqrt{g^2 + f^2 - c}$. Here, $g = -\frac{11}{2}$ and $f = 6$.

Plugging these in:
$$R = \sqrt{\left(-\frac{11}{2}ight)^2 + 6^2 - 5} = \sqrt{\frac{121}{4} + 36 - 5} = \sqrt{\frac{121 + 124}{4}} = \sqrt{\frac{245}{4}} = \frac{\sqrt{245}}{2}$$

Finally, the diameter $D = 2R$ gives us:
$$D = \sqrt{245} = \sqrt{49 	imes 5} = **7\sqrt{5}**$$

Take a moment to appreciate the elegance of this result. We started with an infinite family of circles, used the geometric property of a diameter to constrain our variable, and arrived at a precise, beautiful value.

Animated Solution for Mathematics - Circles: A circle $C$ touches the line $x = 2 y$ at the point $(2, 1)$ and intersects the circle $C_{1} : x^{2} + y^{2} + 2 y - 5 = 0$ at two points $P$ and $Q$ such that $P Q$ is a diameter of $C_{1}$ . Then the diameter of $C$ is :

Select Answer:

Visualized Solution

Visualizing the Tangent Line

Family of Circles Equation

Raw Setup for Circle $C$

Expanding the General Equation

Identifying Circle $C_{1}$

Finding the Common Chord $P Q$

Equation of the Common Chord

Diameter Property of $P Q$

Solving for Parameter $λ$

Final Equation of Circle $C$

Calculating the Radius of $C$

Finding the Diameter of $C$

The Sigma Insight: Director Circle and Family of Circles

Analyzing the Setup

The Intersection

The Power of Subtraction

The Final Reveal

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Select Answer:

Visualized Solution

Visualizing the Tangent Line

Family of Circles Equation

Raw Setup for Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Expanding the General Equation

Identifying Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Finding the Common Chord .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

Equation of the Common Chord

Diameter Property of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ

Solving for Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Final Equation of Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Calculating the Radius of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Finding the Diameter of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C