The line x = y touches a circle at the point (1, 1). If the circle also passes through the point (1,

Q: Wait, why do we use the family of circles formula instead of just finding the center $(h, k)$ directly?

You absolutely could find the center $(h, k)$ by using the fact that the distance from the center to the line equals the radius, and the distance from the center to both points is equal. However, that leads to a system of equations that is much more algebraically intensive. The 'Family of Circles' approach is a powerful shortcut designed specifically for these tangency problems, saving you precious time in the exam.

Q: Why is the line equation $L$ written as $x - y = 0$ and not $x - y = c$?

Great observation! The line is given as $x = y$, which is equivalent to $x - y = 0$. Since the line passes through the origin, the constant term is indeed zero. If the line were $x - y + 5 = 0$, then $L$ would be $x - y + 5$.

Q: Is there any other way to solve this if I forget the family of circles formula?

Yes! You can use the property that the center of the circle lies on the line perpendicular to the tangent at the point of tangency. The slope of the tangent $x = y$ is $1$, so the slope of the normal is $-1$. The normal passes through $(1, 1)$, so its equation is $y - 1 = -1(x - 1)$, or $y = -x + 2$. The center $(h, k)$ must lie on this line, so $k = -h + 2$. Then, use the distance formula: the distance from $(h, k)$ to $(1, 1)$ must equal the distance from $(h, k)$ to $(1, -3)$. This will also give you the correct answer!

Question

The line $x = y$ touches a circle at the point $(1, 1)$. If the circle also passes through the point $(1, -3)$, then its radius is :

Options:

(A) $3\sqrt{2}$

(B) 3

(C) $2\sqrt{2}$

(D) 2

Sigma Percentile · Accepted Answer

### The Geometry of Tangency

Imagine you are standing on a coordinate plane. You see a line, $x = y$, cutting through the origin at a perfect $45^\circ$ angle.

Now, imagine a circle, perfectly balanced, kissing this line at the point $P(1, 1)$. This is not just any circle; it is a member of a vast, infinite family of circles that all share this same point of tangency.

In the world of JEE Advanced, when you see a circle touching a line at a specific point, your mind should immediately jump to the **'Family of Circles'** concept. It is a powerful, elegant tool that allows us to encapsulate an entire infinite set of possibilities into one single, beautiful equation.

### The Magic Formula

To represent any circle that touches the line $L = x - y = 0$ at the point $(1, 1)$, we use the standard form:

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

Why does this work? Think of it as a linear combination. The term $(x - 1)^2 + (y - 1)^2$ represents a point circle at $(1, 1)$, and $\lambda(x - y)$ introduces the tangency constraint.

As we vary $\lambda$, we are essentially 'inflating' or 'deflating' the circle while keeping it anchored to that line at that exact point. It is a brilliant piece of algebraic machinery.

### The Anchor Point

We have our family, but we need our specific circle. The problem gives us a second anchor: the circle must pass through the point $Q(1, -3)$.

This point is our key. If the circle passes through $Q$, then the coordinates $(1, -3)$ must satisfy our equation. Let us substitute these values carefully:

$$(1 - 1)^2 + (-3 - 1)^2 + \lambda(1 - (-3)) = 0$$

Watch the signs here—a single slip can ruin the entire calculation. The first term vanishes to $0$, the second becomes $(-4)^2 = 16$, and the third becomes $\lambda(1 + 3) = 4\lambda$.

We are left with $16 + 4\lambda = 0$, which simplifies beautifully to $\lambda = -4$.

### The Final Reveal

With $\lambda = -4$ in hand, our circle equation is fully defined:

$$(x - 1)^2 + (y - 1)^2 - 4(x - y) = 0$$

Now, we just need to peel back the layers to find the radius. Expanding the squares, we get:

$$(x^2 - 2x + 1) + (y^2 - 2y + 1) - 4x + 4y = 0$$

Combining like terms, we arrive at the general form:

$$x^2 + y^2 - 6x + 2y + 2 = 0$$

Comparing this to the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$, we identify $g = -3$, $f = 1$, and $c = 2$.

The radius formula $r = \sqrt{g^2 + f^2 - c}$ is our final destination. Plugging in our values, we get:

$$r = \sqrt{(-3)^2 + (1)^2 - 2} = \sqrt{9 + 1 - 2} = \sqrt{8}$$

Simplifying $\sqrt{8}$ gives us **$2\sqrt{2}$**. The journey is complete, and the elegance of the result is undeniable.

Animated Solution for Mathematics - Circles: The line $x = y$ touches a circle at the point $(1, 1)$ . If the circle also passes through the point $(1, - 3)$ , then its radius is :

Select Answer:

Visualized Solution

Visualizing the Tangency

Family of Circles Concept

Setting up the Equation

The Second Point Anchor

Substituting Point $Q (1, - 3)$

Solving for $λ$

Updating the Circle Equation

Expanding and Simplifying

Comparing with Standard Form

The Radius Formula

Final Calculation

Conclusion

The Sigma Insight: Director Circle and Family of Circles

The Geometry of Tangency

The Magic Formula

The Anchor Point

The Final Reveal

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Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The line x=y touches a circle at the point (1,1). If the circle also passes through the point (1,−3), then its radius is :