JEE Challenger
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MathHardAdv 2021 • Paper 2

Intersection X Coordinate of Largest Enclosed Circle and Parabola

Comprehension Passage

Let a region be defined as R={(x,y)∈R×R:x≥0 and y2≤4−x}R = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x \ge 0 \text{ and } y^2 \le 4 - x\}. Suppose F\mathcal{F} represents the collection of all circles enclosed within region RR with their centers lying on the xx-axis. Let CC denote the circle possessing the maximum radius among all circles belonging to F\mathcal{F}. Let (α,β)(\alpha, \beta) be a point where the circle CC meets the curve y2=4−xy^2 = 4 - x.

Calculate the value of α\alpha.

Official Numerical Answer2

Topics & Concepts

Conic SectionsCirclesParabolaMaxima and Minima

Step-by-Step Solution

To find the value of α\alpha, we analyze the region RR and the conditions for a circle CC to be completely contained within RR.

1. Definition of the Region and Circle

The region RR is defined by: R={(x,y)∈R×R:x≥0 and y2≤4−x}R = \{(x, y) \in \mathbb{R} \times \mathbb{R} : x \ge 0 \text{ and } y^2 \le 4 - x\}

Let CC be a circle in the family F\mathcal{F} centered at (x0,0)(x_0, 0) with radius r>0r > 0. The equation of the circle is: (x−x0)2+y2=r2(x - x_0)^2 + y^2 = r^2

2. Constraints on the Circle

For CC to be entirely contained within region RR:

  1. x≥0x \ge 0 constraint: The leftmost point on the circle is x=x0−rx = x_0 - r. Therefore, we must have: x0−r≥0  ⟹  r≤x0x_0 - r \ge 0 \implies r \le x_0

  2. y2≤4−xy^2 \le 4 - x constraint: For all points (x,y)(x, y) on the circle, y2=r2−(x−x0)2y^2 = r^2 - (x - x_0)^2. Substituting this into the inequality gives: r2−(x−x0)2≤4−x∀x∈[x0−r,x0+r]r^2 - (x - x_0)^2 \le 4 - x \quad \forall x \in [x_0 - r, x_0 + r]

    Expanding and rearranging the terms: r2−(x2−2x0x+x02)≤4−xr^2 - (x^2 - 2x_0 x + x_0^2) \le 4 - x x2−(2x0+1)x+(x02−r2+4)≥0∀x∈[x0−r,x0+r]x^2 - (2x_0 + 1)x + (x_0^2 - r^2 + 4) \ge 0 \quad \forall x \in [x_0 - r, x_0 + r]

    Define g(x)=x2−(2x0+1)x+(x02−r2+4)g(x) = x^2 - (2x_0 + 1)x + (x_0^2 - r^2 + 4).

3. Maximizing the Radius rr

The vertex of the parabola g(x)g(x) occurs at: xv=2x0+12=x0+12x_v = \frac{2x_0 + 1}{2} = x_0 + \frac{1}{2}

For r≥12r \ge \frac{1}{2}, the vertex xvx_v lies inside the interval [x0−r,x0+r][x_0 - r, x_0 + r]. Hence, the minimum value of g(x)g(x) occurs at x=xvx = x_v. For g(x)≥0g(x) \ge 0 to hold everywhere in the domain: g(x0+12)≥0g\left(x_0 + \frac{1}{2}\right) \ge 0 (x0+12)2−(2x0+1)(x0+12)+x02−r2+4≥0\left(x_0 + \frac{1}{2}\right)^2 - (2x_0 + 1)\left(x_0 + \frac{1}{2}\right) + x_0^2 - r^2 + 4 \ge 0 −(x0+12)2+x02−r2+4≥0-\left(x_0 + \frac{1}{2}\right)^2 + x_0^2 - r^2 + 4 \ge 0 154−x0−r2≥0  ⟹  r2≤154−x0\frac{15}{4} - x_0 - r^2 \ge 0 \implies r^2 \le \frac{15}{4} - x_0

To maximize rr, we combine the two upper bounds on rr:

  1. r≤x0  ⟹  r2≤x02r \le x_0 \implies r^2 \le x_0^2
  2. r2≤154−x0r^2 \le \frac{15}{4} - x_0

The maximum radius occurs when these two bounds are equal: x02=154−x0x_0^2 = \frac{15}{4} - x_0 4x02+4x0−15=04x_0^2 + 4x_0 - 15 = 0 (2x0+5)(2x0−3)=0(2x_0 + 5)(2x_0 - 3) = 0

Since x0>0x_0 > 0, we get x0=32x_0 = \frac{3}{2}, which gives the maximum radius r=32r = \frac{3}{2}.

4. Intersection of Circle CC and Curve y2=4−xy^2 = 4 - x

The equation of the circle CC with center (32,0)\left(\frac{3}{2}, 0\right) and radius r=32r = \frac{3}{2} is: (x−32)2+y2=94\left(x - \frac{3}{2}\right)^2 + y^2 = \frac{9}{4} y2=94−(x2−3x+94)=3x−x2y^2 = \frac{9}{4} - \left(x^2 - 3x + \frac{9}{4}\right) = 3x - x^2

To find the point (α,β)(\alpha, \beta) where circle CC meets the parabola y2=4−xy^2 = 4 - x, set the expressions for y2y^2 equal to each other: 3α−α2=4−α3\alpha - \alpha^2 = 4 - \alpha α2−4α+4=0\alpha^2 - 4\alpha + 4 = 0 (α−2)2=0  ⟹  α=2(\alpha - 2)^2 = 0 \implies \alpha = 2

Thus, the value of α\alpha is 22.

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