Wait, why is the radius of the moving circle equal to $\beta$?

Great question! The circle touches the x-axis, which is the line $y = 0$. The distance from the center $(\alpha, \beta)$ to the line $y = 0$ is simply the absolute value of the y-coordinate, which is $|\beta|$. Since the problem states $\beta > 0$, the radius is exactly $\beta$.

JEE Main 2022 (25 July Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Conic Sections: Let the locus of the centre $(α, β), β > 0$ , of the circle which touches the circle $x^{2} + (y - 1)^{2} = 1$ externally and also touches the x-axis be $L$ . Then the area bounded by $L$ and the line $y = 4$ is :

Q: Can we use integration with respect to $y$ instead of $x$?

Absolutely! You can express $x$ as a function of $y$ ($x = \pm 2\sqrt{y}$) and integrate with respect to $y$ from $0$ to $4$. The area would be $\int_{0}^{4} 2(2\sqrt{y}) dy$. Both methods will yield the same result, $\frac{64}{3}$.

Select Answer:

Visualized Solution

$x^{2} + (y - 1)^{2} = 1$

Given fixed circle: $x^{2} + (y - 1)^{2} = 1$
Center $C_{1} = (0, 1)$
Radius $r_{1} = 1$

$(α, β)$ and $r = β$

Let the moving circle have center $C = (α, β)$ and radius $r$ .
Since it touches the x-axis and $β > 0$ , we have $r = β$ .

$d = r_{1} + r_{2}$

The circles touch externally.
Condition: Distance between centers = Sum of radii.

$(α - 0)^{2} + (β - 1)^{2} = 1 + β$

Substitute the coordinates into the distance formula:
$(α - 0)^{2} + (β - 1)^{2} = 1 + β$

$α^{2} + (β - 1)^{2} = (1 + β)^{2}$

Squaring both sides: $α^{2} + (β - 1)^{2} = (1 + β)^{2}$
Expanding: $α^{2} + β^{2} - 2 β + 1 = 1 + 2 β + β^{2}$

$x^{2} = 4 y$

Simplifying the equation: $α^{2} - 2 β = 2 β \Rightarrow α^{2} = 4 β$
Replacing $(α, β)$ with $(x, y)$ , the locus $L$ is: $x^{2} = 4 y$

$y = 4$

Locus $L : x^{2} = 4 y$ or $y = \frac{x ^{2}}{4}$
Bounding line: $y = 4$
Intersection points: $\frac{x ^{2}}{4} = 4 \Rightarrow x^{2} = 16 \Rightarrow x = \pm 4$

$A_{l e f t} = A_{r i g h t}$

The bounded region is symmetric about the y-axis.
We can calculate the area in the first quadrant and multiply by $2$ .

$A = 2 \int_{0}^{4} (4 - \frac{x ^{2}}{4}) d x$

Area $A = \int_{- 4}^{4} (4 - \frac{x ^{2}}{4}) d x$
By symmetry: $A = 2 \int_{0}^{4} (4 - \frac{x ^{2}}{4}) d x$

$A = 2 [4 x - \frac{x ^{3}}{12}]_{0}^{4}$

Integrating term by term:
$A = 2 [4 x - \frac{x ^{3}}{12}]_{0}^{4}$

$A = \frac{64}{3}$

$A = 2 [(4 (4) - \frac{4 ^{3}}{12}) - 0]$
$A = 2 [16 - \frac{64}{12}] = 2 [16 - \frac{16}{3}]$
$A = 2 [\frac{32}{3}] = \frac{64}{3}$ square units.

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The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Solution Diagram

Analyzing the Setup

Imagine you are standing on a coordinate plane. You see a fixed circle, anchored at

(0, 1)

with a radius of

1

Now, imagine a second, smaller circle that is constantly moving, but it has a strict rule: it must always touch the x-axis and it must always kiss the fixed circle from the outside. This is the geometric dance we are analyzing today.

Defining the Locus

Let the center of our moving circle be

(α, β)

. Because it touches the x-axis and stays above it, its radius

r

must be exactly

β

Now, consider the fixed circle with center

C_{1} = (0, 1)

and radius

r_{1} = 1

. When two circles touch externally, the distance

d

between their centers is the sum of their radii:

d = r_{1} + r_{2}

Here,

r_{2} = β

, so

d = 1 + β

. Using the distance formula, we write:

(α - 0)^{2} + (β - 1)^{2} = 1 + β

Squaring both sides, we get

α^{2} + (β - 1)^{2} = (1 + β)^{2}

. Expanding this, we find:

α^{2} + β^{2} - 2 β + 1 = 1 + 2 β + β^{2}

The

β^{2}

and the

1

cancel out beautifully, leaving us with

α^{2} = 4 β

. Replacing

(α, β)

with

(x, y)

, we find our locus

L

is the parabola

x^{2} = 4 y

Calculating the Area

Now that we have our parabola

y = \frac{x ^{2}}{4}

, we need to find the area bounded by this curve and the line

y = 4

First, find the intersection points:

\frac{x ^{2}}{4} = 4

, which gives

x^{2} = 16

, so

x = \pm 4

. The region is symmetric about the y-axis.

We can calculate the area in the first quadrant and double it. The area

A

is given by:

A = 2 \int_{0}^{4} (4 - \frac{x ^{2}}{4}) d x

Integrating term by term, we get:

2 [4 x - \frac{x ^{3}}{12}]_{0}^{4}

Plugging in the limits, we have:

2 [16 - \frac{64}{12}] = 2 [16 - \frac{16}{3}] = 2 [\frac{32}{3}] = \frac{64}{3}

The final area is $\frac{64}{3}$ square units. It is a beautiful result, isn't it? The complexity of the moving circle simplifies into the elegant geometry of a parabola.

Common Traps & FAQs

Wait, why is the radius of the moving circle equal to

β

Great question! The circle touches the x-axis, which is the line

y = 0

. The distance from the center

(α, β)

to the line

y = 0

is simply the absolute value of the y-coordinate, which is

∣ β ∣

. Since the problem states

β > 0

, the radius is exactly

β

Why do we use the sum of the radii

r_{1} + r_{2}

instead of the difference

∣ r_{1} - r_{2} ∣

The problem explicitly states that the circles touch *externally*. When two circles touch externally, the distance between their centers is the sum of their radii. If they touched internally, the distance would be the difference of their radii.

Can we use integration with respect to

y

instead of

x

Absolutely! You can express

x

as a function of

y

(

x = \pm 2 y

) and integrate with respect to

y

from

0

4

. The area would be

\int_{0}^{4} 2 (2 y) d y

. Both methods will yield the same result,

\frac{64}{3}

Animated Solution for Mathematics - Conic Sections: Let the locus of the centre $(α, β), β > 0$ , of the circle which touches the circle $x^{2} + (y - 1)^{2} = 1$ externally and also touches the x-axis be $L$ . Then the area bounded by $L$ and the line $y = 4$ is :

Select Answer:

Visualized Solution

$x^{2} + (y - 1)^{2} = 1$

$(α, β)$ and $r = β$

$d = r_{1} + r_{2}$

$(α - 0)^{2} + (β - 1)^{2} = 1 + β$

$α^{2} + (β - 1)^{2} = (1 + β)^{2}$

$x^{2} = 4 y$

$y = 4$

$A_{l e f t} = A_{r i g h t}$

$A = 2 \int_{0}^{4} (4 - \frac{x ^{2}}{4}) d x$

$A = 2 [4 x - \frac{x ^{3}}{12}]_{0}^{4}$

$A = \frac{64}{3}$

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

Defining the Locus

Calculating the Area

Similar Questions

Comprehension Passage

If $P$ is any point of $C_{1}$ and $Q$ is another point on $C_{2}$ , then $\frac{P A ^{2} + P B ^{2} + P C ^{2} + P D ^{2}}{Q A ^{2} + Q B ^{2} + Q C ^{2} + Q D ^{2}}$ is equal to

If a circle is such that it touches the line $L$ and the circle $C_{1}$ externally, such that both the circles are on the same side of the line, then the locus of centre of the circle is

A line $L^{'}$ through $A$ is drawn parallel to $B D$ . Point $S$ moves such that its distances from the line $B D$ and the vertex $A$ are equal. If locus of $S$ cuts $L^{'}$ at $T_{2}$ and $T_{3}$ and $A C$ at $T_{1}$ , then area of $Δ T_{1} T_{2} T_{3}$ is

The line $y = x + 1$ meets the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$ at two points $P$ and $Q$ . If $r$ is the radius of the circle with $P Q$ as diameter then $(3 r)^{2}$ is equal to

If the points of intersection of two distinct conics $x^{2} + y^{2} = 4 b$ and $\frac{x ^{2}}{16} + \frac{y ^{2}}{b ^{2}} = 1$ lie on the curve $y^{2} = 3 x^{2}$ , then $33$ times the area of the rectangle formed by the intersection points is ______

Let $O (0, 0)$ and $A (0, 1)$ be two fixed points. Then the locus of a point $P$ such that the perimeter of $Δ A O P$ is 4, is :

Let $S_{1} : x^{2} + y^{2} = 9$ and $S_{2} : (x - 2)^{2} + y^{2} = 1$ . Then the locus of center of a variable circle $S$ which touches $S_{1}$ internally and $S_{2}$ externally always passes through the points:

The locus of a point which divides the line segment joining the point $(0, - 1)$ and a point on the parabola, $x^{2} = 4 y$ , internally in the ratio 1: 2, is :

If the line $a x + y = c$ , touches both the curves $x^{2} + y^{2} = 1$ and $y^{2} = 42 x$ , then $∣ c ∣$ is equal to :

From a point $A$ common tangents are drawn to the circle $x^{2} + y^{2} = a^{2} /2$ and parabola $y^{2} = 4 a x$ . Find the area of the quadrilateral formed by the common tangents, the chord of contact of the circle and the chord of contact of the parabola.

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2+(y−1)2=1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (α,β) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r=β

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d=r1​+r2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (α−0)2+(β−1)2​=1+β

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2+(β−1)2=(1+β)2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2=4y

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=4

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Aleft​=Aright​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A=2∫04​(4−4x2​)dx

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A=2[4x−12x3​]04​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A=364​

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

Defining the Locus

Calculating the Area

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If P is any point of C1​ and Q is another point on C2​, then QA2+QB2+QC2+QD2PA2+PB2+PC2+PD2​ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The line y=x+1 meets the ellipse 4x2​+2y2​=1 at two points P and Q. If r is the radius of the circle with PQ as diameter then (3r)2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let O(0,0) and A(0,1) be two fixed points. Then the locus of a point P such that the perimeter of ΔAOP is 4, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of a point which divides the line segment joining the point (0,−1) and a point on the parabola, x2=4y, internally in the ratio 1: 2, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line ax+y=c, touches both the curves x2+y2=1 and y2=42​x, then ∣c∣ is equal to :

$x^{2} + (y - 1)^{2} = 1$

$(α, β)$ and $r = β$

$d = r_{1} + r_{2}$

$(α - 0)^{2} + (β - 1)^{2} = 1 + β$

$α^{2} + (β - 1)^{2} = (1 + β)^{2}$

$x^{2} = 4 y$

$y = 4$

$A_{l e f t} = A_{r i g h t}$

$A = 2 \int_{0}^{4} (4 - \frac{x ^{2}}{4}) d x$

$A = 2 [4 x - \frac{x ^{3}}{12}]_{0}^{4}$

$A = \frac{64}{3}$

If $P$ is any point of $C_{1}$ and $Q$ is another point on $C_{2}$ , then $\frac{P A ^{2} + P B ^{2} + P C ^{2} + P D ^{2}}{Q A ^{2} + Q B ^{2} + Q C ^{2} + Q D ^{2}}$ is equal to

The line $y = x + 1$ meets the ellipse $\frac{x ^{2}}{4} + \frac{y ^{2}}{2} = 1$ at two points $P$ and $Q$ . If $r$ is the radius of the circle with $P Q$ as diameter then $(3 r)^{2}$ is equal to

Let $O (0, 0)$ and $A (0, 1)$ be two fixed points. Then the locus of a point $P$ such that the perimeter of $Δ A O P$ is 4, is :

The locus of a point which divides the line segment joining the point $(0, - 1)$ and a point on the parabola, $x^{2} = 4 y$ , internally in the ratio 1: 2, is :

If the line $a x + y = c$ , touches both the curves $x^{2} + y^{2} = 1$ and $y^{2} = 42 x$ , then $∣ c ∣$ is equal to :