JEE Main 2005

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Animated Solution for Mathematics - Conic Sections: A circle touches the x- axis and also touches the circle with centre at $(0, 3)$ and radius 2. The locus of the centre of the circle is

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Visualized Solution

Visualizing the Setup

Fixed circle: Center $C_{1} (0, 3)$ , Radius $r_{1} = 2$
Variable circle touches the x-axis and $C_{1}$ externally.
Goal: Find the locus of the center $(h, k)$ of the variable circle.

Defining the Variable Circle

Let the center of the required circle be $P (h, k)$ .
Let its radius be $r$ .

Condition 1: Touching the $x$ -axis

Since the circle touches the x-axis, its radius $r$ is equal to the distance from $P (h, k)$ to the x-axis.
Therefore, $r = ∣ k ∣$ . Assuming $k > 0$ , we have $r = k$ .

Condition 2: Touching the Fixed Circle

The distance between centers $d$ must equal the sum of radii $r + r_{1}$ .
Fixed circle center $C_{1} (0, 3)$ , radius $r_{1} = 2$ .
Distance $d = (h - 0)^{2} + (k - 3)^{2}$ .

Setting up the Equation

Condition: $h^{2} + (k - 3)^{2} = r + 2$
Substitute $r = k$ :
$h^{2} + (k - 3)^{2} = k + 2$

Squaring Both Sides

Square both sides to eliminate the radical:
$h^{2} + (k - 3)^{2} = (k + 2)^{2}$

Expanding the Terms

Expand $(k - 3)^{2}$ and $(k + 2)^{2}$ :
$h^{2} + k^{2} - 6 k + 9 = k^{2} + 4 k + 4$

Simplifying the Equation

Subtract $k^{2}$ from both sides:
$h^{2} - 6 k + 9 = 4 k + 4$
Rearrange terms:
$h^{2} = 4 k + 6 k + 4 - 9$
Result: $h^{2} = 10 k - 5$

Identifying the Locus

Replace $(h, k)$ with $(x, y)$ for the locus:
$x^{2} = 10 y - 5$
This can be written as: $x^{2} = 10 (y - \frac{1}{2})$

Conclusion and Takeaway

The equation $x^{2} = 10 (y - \frac{1}{2})$ is of the form $X^{2} = 4 aY$ .
This represents a parabola opening upwards.
Therefore, the locus of the center of the circle is a parabola.

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

Solution Diagram

Analyzing the Setup

We are tasked with finding the locus of the center

P (h, k)

of a variable circle with radius

r

that satisfies two conditions: it is tangent to the

x

-axis and externally tangent to a fixed circle centered at

(0, 3)

with radius

2

Since the variable circle is tangent to the

x

-axis (

y = 0

), the distance from its center

(h, k)

to the

x

-axis must equal its radius. Thus, we establish the relationship:

r = k

The Master Equation

When two circles are tangent externally, the distance between their centers is equal to the sum of their radii. The fixed circle has center

C_{1} (0, 3)

and radius

r_{1} = 2

The distance between

P (h, k)

and

C_{1} (0, 3)

is given by the distance formula:

(h - 0)^{2} + (k - 3)^{2}

Setting this distance equal to the sum of the radii (

r + r_{1} = k + 2

), we obtain the master equation:

h^{2} + (k - 3)^{2} = k + 2

Simplifying the Geometry

To eliminate the radical, we square both sides of the equation:

h^{2} + (k - 3)^{2} = (k + 2)^{2}

Expanding the squared binomials on both sides yields:

h^{2} + k^{2} - 6 k + 9 = k^{2} + 4 k + 4

We observe that the

k^{2}

terms cancel out from both sides of the equation, simplifying the expression significantly:

h^{2} - 6 k + 9 = 4 k + 4

Final Calculation

Rearranging the terms to isolate

h^{2}

on one side, we get:

h^{2} = 10 k - 5

Factoring the right side, we arrive at the standard form:

h^{2} = 10 (k - \frac{1}{2})

Replacing

(h, k)

with the general coordinates

(x, y)

, the locus of the center is:

x^{2} = 10 (y - \frac{1}{2})

This equation represents a parabola that opens upwards with its vertex at

(0, \frac{1}{2})

Common Traps & FAQs

Why is the radius

r

equal to

k

and not

h

? I always get confused about which coordinate represents the distance to the x-axis.

That is a classic point of confusion! Think about the geometry: the x-axis is a horizontal line defined by the equation

y = 0

. The distance from any point

(h, k)

to a horizontal line is simply the vertical separation, which is the absolute value of the y-coordinate,

∣ k ∣

. Since our circle is in the upper half-plane,

r = k

. If it were touching the y-axis, the distance would be the horizontal separation,

∣ h ∣

How do I know for sure that it is external tangency and not internal?

Great question! The problem states the circle 'touches' the fixed circle. In the context of locus problems involving a moving circle and a fixed one, unless specified as 'enclosing' or 'internal', we assume external tangency. Geometrically, if the variable circle were inside the fixed one, the distance between centers would be

∣ r_{1} - r ∣

, which would lead to a different locus. Always look for keywords like 'externally' or 'internally' in the problem statement!

Why does the

k^{2}

term cancel out? Does that mean I made a mistake?

Not at all! That cancellation is actually the 'aha!' moment of the problem. When the squared terms cancel, it signals that the equation is no longer a circle (which would require both

h^{2}

and

k^{2}

terms to persist). The fact that

k^{2}

vanishes leaves us with a linear term in

k

and a squared term in

h

, which is the hallmark of a parabola. Embrace the cancellation; it is the math telling you the shape of the curve!

Animated Solution for Mathematics - Conic Sections: A circle touches the x- axis and also touches the circle with centre at $(0, 3)$ and radius 2. The locus of the centre of the circle is

Select Answer:

Visualized Solution

Visualizing the Setup

Defining the Variable Circle

Condition 1: Touching the $x$ -axis

Condition 2: Touching the Fixed Circle

Setting up the Equation

Squaring Both Sides

Expanding the Terms

Simplifying the Equation

Identifying the Locus

Conclusion and Takeaway

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

Analyzing the Setup

The Master Equation

Simplifying the Geometry

Final Calculation

Similar Questions

The locus of a variable point whose distance from $(2, 0)$ is $2/3$ times its distance from the line $x = - 9/2$ is

Let $C_{1}$ and $C_{2}$ be two circles with $C_{2}$ lying inside $C_{1}$ . A circle $C$ lying inside $C_{1}$ touches $C_{1}$ internally and $C_{2}$ externally. Identify the locus of the centre of $C$ .

The locus of the point of intersection of the lines, $2 x - y + 42 k = 0$ and $2 k x + k y - 42 = 0$ (k is any non-zero real parameter), 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:

Match the conics in Column I with the statements/expressions in Column II.

List-I

List-II

The equation $\frac{x ^{2}}{1 - r} - \frac{y ^{2}}{1 + r} = 1, r > 1$ represents

The equation $2 x^{2} + 3 y^{2} - 8 x - 18 y + 35 = k$ represents

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

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 :

Let $(x, y)$ be any point on the parabola $y^{2} = 4 x$ . Let $P$ be the point that divides the line segment from $(0, 0)$ to $(x, y)$ in the ratio $1 : 3$ . Then the locus of $P$ is

Select Answer:

Visualized Solution

Visualizing the Setup

Defining the Variable Circle

Condition 1: Touching the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-axis

Condition 2: Touching the Fixed Circle

Setting up the Equation

Squaring Both Sides

Expanding the Terms

Simplifying the Equation

Identifying the Locus

Conclusion and Takeaway

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

Analyzing the Setup

The Master Equation

Simplifying the Geometry

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of a variable point whose distance from (2,0) is 2/3 times its distance from the line x=−9/2 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the point of intersection of the lines, 2​x−y+42​k=0 and 2​kx+ky−42​=0 (k is any non-zero real parameter), is :

Match the conics in Column I with the statements/expressions in Column II.

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation 1−rx2​−1+ry2​=1,r>1 represents

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation 2x2+3y2−8x−18y+35=k represents

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; } Let (x,y) be any point on the parabola y2=4x. Let P be the point that divides the line segment from (0,0) to (x,y) in the ratio 1:3. Then the locus of P is

Condition 1: Touching the $x$ -axis

The locus of a variable point whose distance from $(2, 0)$ is $2/3$ times its distance from the line $x = - 9/2$ is

The locus of the point of intersection of the lines, $2 x - y + 42 k = 0$ and $2 k x + k y - 42 = 0$ (k is any non-zero real parameter), is :

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The equation $2 x^{2} + 3 y^{2} - 8 x - 18 y + 35 = k$ represents

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

Let $(x, y)$ be any point on the parabola $y^{2} = 4 x$ . Let $P$ be the point that divides the line segment from $(0, 0)$ to $(x, y)$ in the ratio $1 : 3$ . Then the locus of $P$ is