The radius of a circle, having minimum area, which touches the curve y = 4 – x² and the lines, y = |

Q: Why does the circle have to touch the vertex $(0, 4)$? Couldn't it touch the parabola at two other points?

That is a fantastic observation! While a circle *can* touch a parabola at two points, we are looking for the circle with the *minimum area* that is constrained by the V-shaped lines $y = |x|$. Because the entire system is symmetric about the $y$-axis, the smallest circle that fits snugly in this region will be pinned against the vertex of the parabola. If it touched elsewhere, it would have to be larger to span the gap between the lines, thus increasing its area.

Q: Why is the center $(0, \beta)$? Can't the center be somewhere else?

Great question. The center must lie on the $y$-axis because the entire region is symmetric about the $y$-axis. If the center were at $(h, k)$ where $h \neq 0$, the circle would not be equidistant from the lines $y = x$ and $y = -x$, meaning it would not be able to touch both lines simultaneously while maintaining the same radius. Symmetry is a powerful tool in JEE problems—use it to simplify your variables!

Q: How do I know the distance from $(0, \beta)$ to $x - y = 0$ is $\frac{\beta}{\sqrt{2}}$?

Let's apply the perpendicular distance formula $d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$. Here, the line is $1x - 1y + 0 = 0$, so $a=1, b=-1, c=0$. The point is $(0, \beta)$. Plugging these in: $d = \frac{|1(0) - 1(\beta) + 0|}{\sqrt{1^2 + (-1)^2}} = \frac{|-\beta|}{\sqrt{1 + 1}} = \frac{\beta}{\sqrt{2}}$. Since the center is above the origin, $\beta$ is positive, so we can drop the absolute value bars.

Question

The radius of a circle, having minimum area, which touches the curve $y = 4 - x^2$ and the lines, $y = |x|$ is:

Options:

(A) $2(\sqrt{2} + 1)$

(B) $2(\sqrt{2} - 1)$

(C) $4(\sqrt{2} - 1)$

(D) $4(\sqrt{2} + 1)$

Sigma Percentile · Accepted Answer

### The Geometry of Balance: A Journey into Symmetry

Imagine standing on the Cartesian plane, looking at a beautiful, downward-opening parabola defined by $y = 4 - x^2$. It stands like a mountain peak, its vertex resting proudly at $(0, 4)$.

Below it, a V-shaped valley formed by the lines $y = |x|$—or more precisely, the lines $y = x$ and $y = -x$—creates a perfect, symmetric cradle. Our mission is to find the radius of the smallest circle that can nestle perfectly within this space, touching both the lines and the parabola.

This is not just a math problem; it is a study in geometric harmony.

### Phase 1: The Y-Axis Anchor

Before we dive into the algebra, let's pause and observe. The parabola is symmetric about the $y$-axis, and the V-shaped lines are also symmetric about the $y$-axis.

In physics and geometry, whenever you see such perfect symmetry, you should immediately suspect that the solution lies along the axis of symmetry. If our circle is to be the "minimum area" circle, it must be perfectly centered.

Therefore, the center of our circle must lie on the $y$-axis. Let us define the center as $C(0, \beta)$ and the radius as $r$. This simple decision reduces our variables significantly, turning a complex 2D problem into a manageable one-dimensional search for $\beta$ and $r$.

### Phase 2: The Line Tangency

Now, let's focus on the lines. For the circle to touch the line $y = x$ (which we can rewrite as $x - y = 0$), the perpendicular distance from the center $C(0, \beta)$ to this line must be exactly equal to the radius $r$.

We invoke the classic distance formula for a point $(x_1, y_1)$ to a line $ax + by + c = 0$:

$$d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$$

Substituting our values, we get:

$$r = \frac{|1(0) - 1(\beta)|}{\sqrt{1^2 + (-1)^2}} = \frac{|-\beta|}{\sqrt{2}}$$

Since the circle is in the upper region, $\beta$ is positive, so we simplify this to $r = \frac{\beta}{\sqrt{2}}$, or more elegantly, $\beta = r\sqrt{2}$. This is our first crucial relationship. We have locked the center's height to the radius.

### Phase 3: The Vertex Connection

Next, we must address the parabola. The circle is wedged between the lines and the parabola.

Because of the symmetry we discussed, the circle will touch the parabola at its highest point, the vertex $(0, 4)$. The distance from the center $C(0, \beta)$ to the vertex $(0, 4)$ is simply the radius $r$.

Since the center is below the vertex, the vertical distance is simply $r = 4 - \beta$.

### Phase 4: The Algebraic Victory

We now have a system of two equations with two variables:

1. $\beta = r\sqrt{2}$
2. $r = 4 - \beta$

Let's substitute the first into the second:

$$r = 4 - r\sqrt{2}$$

Now, we perform the algebraic dance. Move the $r$ terms to one side:

$$r + r\sqrt{2} = 4$$

$$r(1 + \sqrt{2}) = 4$$

$$r = \frac{4}{\sqrt{2} + 1}$$

To reach the final answer, we rationalize the denominator by multiplying by the conjugate, $(\sqrt{2} - 1)$:

$$r = \frac{4(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)}$$

The denominator becomes $(\sqrt{2})^2 - 1^2 = 2 - 1 = 1$. Thus, we arrive at our destination:

**$$r = 4(\sqrt{2} - 1)$$**

### Conclusion

Look at what we have achieved. By trusting the symmetry of the geometry, we transformed a daunting problem into a clean, elegant algebraic solution.

The radius is **$4(\sqrt{2} - 1)$**. Remember, in JEE Advanced, the math is rarely about brute force; it is about finding the path of least resistance through the geometry.

Keep visualizing, keep simplifying, and the answers will always reveal themselves.

Animated Solution for Mathematics - Circles: The radius of a circle, having minimum area, which touches the curve $y = 4 - x^{2}$ and the lines, $y = ∣ x ∣$ is:

Select Answer:

Visualized Solution

Visualizing the Geometry

Symmetry and the Circle's Center

Tangency with $y = ∣ x ∣$

Setting up the Distance Equation

Simplifying the Relation

Tangency with the Parabola

Radius from the Vertex

Substituting $β$

Solving for $r$

Isolating the Radius

Rationalizing the Denominator

Final Calculation

The Sigma Insight: Equation of Tangent and Normal

The Geometry of Balance

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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Comprehension Passage

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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 radius of a circle, having minimum area, which touches the curve y=4−x2 and the lines, y=∣x∣ is: