JEE Main 2014

LEVELJEE Main

Animated Solution for Mathematics - Circles: Let $C$ be the circle with centre at $(1, 1)$ and radius $= 1$ . If $T$ is the circle centred at $(0, y)$ , passing through origin and touching the circle $C$ externally, then the radius of $T$ is equal to

Q: Wait, why is the radius of circle $T$ equal to $|y|$?

The center of circle $T$ is at $(0, y)$ and it passes through the origin $(0, 0)$. The radius is the distance between these two points. Using the distance formula, $r = \sqrt{(0 - 0)^2 + (y - 0)^2} = \sqrt{y^2} = |y|$. We use the absolute value because a radius must always be a positive quantity.

Q: Why does the external touch condition $d = R + r$ hold?

This is a fundamental property of geometry. When two circles touch externally, the shortest path between their centers passes through the point of contact, and this path is exactly the sum of their radii. If they were touching internally, the distance would be the difference of their radii.

Q: Why did the $y < 0$ case lead to a contradiction?

When we assumed $y < 0$, the absolute value $|y|$ became $-y$. Substituting this into our equation $2 - 2y = 1 + 2|y|$ gave us $2 - 2y = 1 - 2y$. The $-2y$ terms cancel out, leaving $2 = 1$, which is a mathematical impossibility. This tells us that no circle centered at $(0, y)$ with $y < 0$ can touch circle $C$ externally while passing through the origin.

Select Answer:

Visualized Solution

Defining Circle $C$

Circle $C$ has its center at $(1, 1)$ .
The radius of circle $C$ is given as $R = 1$ .

Defining Circle $T$

Circle $T$ has its center on the y-axis at $(0, y)$ .
It passes through the origin $(0, 0)$ .

Radius of Circle $T$

Since $T$ passes through $(0, 0)$ , its radius $r$ is the distance from its center $(0, y)$ to the origin.
$r = (0 - 0)^{2} + (y - 0)^{2} = ∣ y ∣$

The External Touching Condition

The two circles $C$ and $T$ touch each other externally.
Condition for external touch: Distance between centers $d = R + r$

Setting up the Distance Equation

Distance between centers $(1, 1)$ and $(0, y)$ is $(1 - 0)^{2} + (1 - y)^{2}$
Sum of radii is $1 + ∣ y ∣$
Equation: $1 + (1 - y)^{2} = 1 + ∣ y ∣$

Squaring Both Sides

To remove the square root, square both sides of the equation.
$(1 + (1 - y)^{2})^{2} = (1 + ∣ y ∣)^{2}$
$1 + (1 - y)^{2} = (1 + ∣ y ∣)^{2}$

Expanding the Terms

Expand $(1 - y)^{2}$ : $1 - 2 y + y^{2}$
Expand $(1 + ∣ y ∣)^{2}$ : $1 + 2∣ y ∣ + ∣ y ∣^{2}$
Note that $∣ y ∣^{2} = y^{2}$
$1 + (1 - 2 y + y^{2}) = 1 + 2∣ y ∣ + y^{2}$

Algebraic Simplification

Equation: $2 - 2 y + y^{2} = 1 + 2∣ y ∣ + y^{2}$
Cancel $y^{2}$ from both sides.
Simplified: $2 - 2 y = 1 + 2∣ y ∣$

Solving for $y$ (Case 1: $y > 0$ )

Assume $y > 0$ , so $∣ y ∣ = y$
Substitute into equation: $2 - 2 y = 1 + 2 y$
Rearrange: $4 y = 1$
$y = \frac{1}{4}$

Checking Case 2 ( $y < 0$ )

Assume $y < 0$ , so $∣ y ∣ = - y$
Substitute into equation: $2 - 2 y = 1 - 2 y$
Simplify: $2 = 1$
This is a contradiction, so $y$ cannot be negative.

Final Result

The only valid solution is $y = \frac{1}{4}$
The radius of circle $T$ is $r = ∣ y ∣ = \frac{1}{4}$
Final Answer: $\frac{1}{4}$

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The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

Circle

C

is centered at

O_{C} = (1, 1)

with a radius

R = 1

Circle

T

has its center on the y-axis at

O_{T} = (0, y)

. Since it passes through the origin

(0, 0)

, its radius

r

is the distance from

(0, y)

(0, 0)

, which is

r = ∣ y ∣

The Tangency Condition

For two circles to touch externally, the distance

d

between their centers must equal the sum of their radii:

d = R + r

The distance

d

between

(1, 1)

and

(0, y)

is calculated using the distance formula:

d = (1 - 0)^{2} + (1 - y)^{2} = 1 + (1 - y)^{2}

Equating this to the sum of the radii

1 + ∣ y ∣

, we obtain the master equation:

1 + (1 - y)^{2} = 1 + ∣ y ∣

The Algebraic Dance

To solve for

y

, we square both sides of the equation:

1 + (1 - y)^{2} = (1 + ∣ y ∣)^{2}

Expanding both sides yields:

1 + (1 - 2 y + y^{2}) = 1 + 2∣ y ∣ + ∣ y ∣^{2}

Since

∣ y ∣^{2} = y^{2}

, the

y^{2}

terms cancel out, simplifying the expression to:

2 - 2 y = 1 + 2∣ y ∣

Final Calculation

We now evaluate the two possible cases for the absolute value:

Case 1: $y > 0$ Here,

∣ y ∣ = y

. The equation becomes:

2 - 2 y = 1 + 2 y

4 y = 1

y = \frac{1}{4}

Case 2: $y < 0$ Here,

∣ y ∣ = - y

. The equation becomes:

2 - 2 y = 1 - 2 y

2 = 1

This is a contradiction, meaning no solution exists for

y < 0

The only valid solution is

y = \frac{1}{4}

. Therefore, the radius of circle

T

is $r = \frac{1}{4}$ .

Common Traps & FAQs

Wait, why is the radius of circle

T

equal to

∣ y ∣

The center of circle

T

is at

(0, y)

and it passes through the origin

(0, 0)

. The radius is the distance between these two points. Using the distance formula,

r = (0 - 0)^{2} + (y - 0)^{2} = y^{2} = ∣ y ∣

. We use the absolute value because a radius must always be a positive quantity.

Why does the external touch condition

d = R + r

hold?

This is a fundamental property of geometry. When two circles touch externally, the shortest path between their centers passes through the point of contact, and this path is exactly the sum of their radii. If they were touching internally, the distance would be the difference of their radii.

Why did the

y < 0

case lead to a contradiction?

When we assumed

y < 0

, the absolute value

∣ y ∣

became

- y

. Substituting this into our equation

2 - 2 y = 1 + 2∣ y ∣

gave us

2 - 2 y = 1 - 2 y

. The

- 2 y

terms cancel out, leaving

2 = 1

, which is a mathematical impossibility. This tells us that no circle centered at

(0, y)

with

y < 0

can touch circle

C

externally while passing through the origin.

Animated Solution for Mathematics - Circles: Let $C$ be the circle with centre at $(1, 1)$ and radius $= 1$ . If $T$ is the circle centred at $(0, y)$ , passing through origin and touching the circle $C$ externally, then the radius of $T$ is equal to

Select Answer:

Visualized Solution

Defining Circle $C$

Defining Circle $T$

Radius of Circle $T$

The External Touching Condition

Setting up the Distance Equation

Squaring Both Sides

Expanding the Terms

Algebraic Simplification

Solving for $y$ (Case 1: $y > 0$ )

Checking Case 2 ( $y < 0$ )

Final Result

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Tangency Condition

The Algebraic Dance

Final Calculation

Similar Questions

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The locus of the centres of the circles, which touch the circle, $x^{2} + y^{2} = 1$ externally, also touch the y-axis and lie in the first quadrant, is :

The centre of a circle passing through the points $(0, 0), (1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ is

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The circle passing through the point $(- 1, 0)$ and touching the y-axis at $(0, 2)$ also passes through the point

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Select Answer:

Visualized Solution

Defining Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Defining Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Radius of Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

The External Touching Condition

Setting up the Distance Equation

Squaring Both Sides

Expanding the Terms

Algebraic Simplification

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y (Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y>0)

Checking Case 2 ( .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y<0)

Final Result

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Tangency Condition

The Algebraic Dance

Final Calculation

Similar Questions

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Defining Circle $C$

Defining Circle $T$

Radius of Circle $T$

Solving for $y$ (Case 1: $y > 0$ )

Checking Case 2 ( $y < 0$ )

A circle is given by $x^{2} + (y - 1)^{2} = 1$ , another circle $C$ touches it externally and also the x-axis, then the locus of its centre is

The locus of the centres of the circles, which touch the circle, $x^{2} + y^{2} = 1$ externally, also touch the y-axis and lie in the first quadrant, is :

The centre of a circle passing through the points $(0, 0), (1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ is

The centre of the circle passing through $(0, 0)$ and $(1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ is

A circle passes through the points (2,3) and (4,5). If its centre lies on the line, $y - 4 x + 3 = 0$ , then its radius is equal to

The circle passing through the point $(- 1, 0)$ and touching the y-axis at $(0, 2)$ also passes through the point

If one of the diameters o of the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is a chord of another circle 'C' whose center is at (2, 1), then its radius is

If one of the diameters of the circle $x^{2} + y^{2} - 2 x - 6 y + 6 = 0$ is a chord to the circle with centre $(2, 1)$ , then the radius of the circle is