JEE Main 2018 (15 April Evening)

LEVELJEE Main

Animated Solution for Mathematics - Circles: The tangent to the circle $C_{1} : x^{2} + y^{2} - 2 x - 1 = 0$ at the point (2, 1) cuts off a chord of length 4 from a circle $C_{2}$ whose centre is (3, -2). The radius of $C_{2}$ is :-

Q: I got $R^2 = 6$, but the options are $\sqrt{2}, \sqrt{6}, 3, 2$. Did I miss a step?

You are doing fantastic! You have successfully calculated $R^2$, which is the square of the radius. The question asks for the radius $R$ itself. Just take the square root of your result, and you will arrive at $\sqrt{6}$. You are right on the finish line!

Select Answer:

Visualized Solution

Identify Circle $C_{1}$ and Point $P$

Given circle $C_{1} : x^{2} + y^{2} - 2 x - 1 = 0$
Point $P (2, 1)$ lies on $C_{1}$ because $2^{2} + 1^{2} - 2 (2) - 1 = 0$

Equation of Tangent at a Point

The equation of a tangent to a circle at a point $(x_{1}, y_{1})$ is given by $T = 0$ .
Transformations: $x^{2} \to x x_{1}$ , $y^{2} \to y y_{1}$ , $2 x \to x + x_{1}$

Substitute $P (2, 1)$ into $T = 0$

Substitute $(x_{1}, y_{1}) = (2, 1)$ into the transformed equation.
$x (2) + y (1) - (x + 2) - 1 = 0$

Simplify to get Tangent Equation

$2 x + y - x - 2 - 1 = 0$
$x + y - 3 = 0$

Introduce Circle $C_{2}$ and the Chord

A second circle $C_{2}$ has its center at $O_{2} (3, - 2)$ .
The tangent line $x + y - 3 = 0$ acts as a chord for $C_{2}$ .
The length of this chord is given as $4$ .

Perpendicular Distance Formula

The perpendicular distance $p$ from a point $(x_{1}, y_{1})$ to a line $a x + b y + c = 0$ is:
$p = \frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}$

Substitute Values for Distance $p$

Center $O_{2} (3, - 2)$ and line $x + y - 3 = 0$ .
$p = \frac{∣ ( 1 ) ( 3 ) + ( 1 ) ( - 2 ) - 3∣}{1 ^{2} + 1 ^{2}}$

Calculate Perpendicular Distance $p$

$p = \frac{∣3 - 2 - 3∣}{1 + 1}$
$p = \frac{∣ - 2∣}{2} = \frac{2}{2} = 2$

Pythagoras Theorem in the Circle

A perpendicular from the center bisects the chord.
Half-chord length $= \frac{4}{2} = 2$
By Pythagoras theorem: $R^{2} = p^{2} + (half-chord)^{2}$

Substitute into Pythagoras Theorem

Substitute $p = 2$ and half-chord $= 2$ .
$R^{2} = (2)^{2} + (2)^{2}$

Calculate $R^{2}$

$R^{2} = 2 + 4$
$R^{2} = 6$

Final Answer

Since $R^{2} = 6$ , taking the square root gives $R = 6$ .
The radius of circle $C_{2}$ is $6$ .

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We begin with the circle

C_{1} : x^{2} + y^{2} - 2 x - 1 = 0

. We are given a point

P (2, 1)

on this circle.

First, we verify that

P

lies on the circle by substituting

x = 2

and

y = 1

2^{2} + 1^{2} - 2 (2) - 1 = 4 + 1 - 4 - 1 = 0

The point satisfies the equation perfectly.

The Tangent as a Bridge

To find the tangent at

P (2, 1)

, we utilize the elegant

T = 0

transformation. By replacing

x^{2}

with

x x_{1}

y^{2}

with

y y_{1}

, and

2 x

with

(x + x_{1})

, we transform the circle equation into the tangent line equation.

Substituting

(x_{1}, y_{1}) = (2, 1)

into the transformation:

x (2) + y (1) - (x + 2) - 1 = 0

Simplifying this expression:

2 x + y - x - 2 - 1 = 0

x + y - 3 = 0

This line serves as our bridge between the two geometric entities.

The Geometry of the Chord

Now, consider circle

C_{2}

with center

O_{2} (3, - 2)

. The line

x + y - 3 = 0

acts as a chord of length

4

within this circle.

To find the radius

R

C_{2}

, we construct a right-angled triangle. We drop a perpendicular of length

p

from the center

O_{2} (3, - 2)

to the chord.

A fundamental property of circles dictates that this perpendicular bisects the chord. Since the total length of the chord is

4

, the half-chord length is

2

By the Pythagorean theorem, the relationship between the radius

R

, the perpendicular distance

p

, and the half-chord is:

R^{2} = p^{2} + 2^{2}

Final Calculation

We calculate the perpendicular distance

p

from the center

(3, - 2)

to the line

x + y - 3 = 0

using the formula

p = \frac{∣ a x _{0} + b y _{0} + c ∣}{a ^{2} + b ^{2}}

p = \frac{∣ ( 1 ) ( 3 ) + ( 1 ) ( - 2 ) - 3∣}{1 ^{2} + 1 ^{2}}

Simplifying the numerator and denominator:

p = \frac{∣3 - 2 - 3∣}{2} = \frac{∣ - 2∣}{2} = \frac{2}{2} = 2

Finally, we substitute

p = 2

into our Pythagorean equation:

R^{2} = (2)^{2} + 2^{2}

R^{2} = 2 + 4 = 6

The radius of the circle

C_{2}

is $R = 6$ .

Common Traps & FAQs

Wait, why do we use the

T = 0

method for the tangent? Can't we just find the slope and use point-slope form?

You absolutely can! Finding the slope of the radius and then the negative reciprocal for the tangent is a perfectly valid path. However, the

T = 0

method is the 'JEE shortcut'—it's elegant, fast, and significantly reduces the chance of calculation errors. Think of it as a master key that unlocks the equation of the tangent instantly for any conic section.

Why does the perpendicular from the center bisect the chord?

Think of the circle as a perfectly symmetric object. If you drop a perpendicular from the center to a chord, you are essentially creating two congruent right-angled triangles. Because the hypotenuse is the radius in both cases and the perpendicular side is shared, the remaining sides must be equal. It is a fundamental property of symmetry that makes our life much easier!

I got

R^{2} = 6

, but the options are

2, 6, 3, 2

. Did I miss a step?

You are doing fantastic! You have successfully calculated

R^{2}

, which is the square of the radius. The question asks for the radius

R

itself. Just take the square root of your result, and you will arrive at

6

. You are right on the finish line!

Animated Solution for Mathematics - Circles: The tangent to the circle $C_{1} : x^{2} + y^{2} - 2 x - 1 = 0$ at the point (2, 1) cuts off a chord of length 4 from a circle $C_{2}$ whose centre is (3, -2). The radius of $C_{2}$ is :-

Select Answer:

Visualized Solution

Identify Circle $C_{1}$ and Point $P$

Equation of Tangent at a Point

Substitute $P (2, 1)$ into $T = 0$

Simplify to get Tangent Equation

Introduce Circle $C_{2}$ and the Chord

Perpendicular Distance Formula

Substitute Values for Distance $p$

Calculate Perpendicular Distance $p$

Pythagoras Theorem in the Circle

Substitute into Pythagoras Theorem

Calculate $R^{2}$

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent as a Bridge

The Geometry of the Chord

Final Calculation

Similar Questions

The line $2 x - y + 1 = 0$ is a tangent to the circle at the point $(2, 5)$ and the centre of the circle lies on $x - 2 y = 4$ . Then, the radius of the circle is:

If a circle C passing through the point (4,0) touches the circle $x^{2} + y^{2} + 4 x - 6 y = 12$ externally at the point (1, -1), then the radius of C is :

Let $T_{1}, T_{2}$ be two tangents drawn from $(- 2, 0)$ onto the circle $C : x^{2} + y^{2} = 1$ . Determine the circles touching $C$ and $T_{1}, T_{2}$ as their pair of tangents. Further, find the equations of all possible common tangents to these circles, when taken two at a time.

Two circles with equal radii are intersecting at the points $(0, 1)$ and $(0, - 1)$ . The tangent at the point $(0, 1)$ to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is :

Two circles, each of radius 5 units, touch each other at $(1, 2)$ . If the equation of their common tangent is $4 x + 3 y = 10$ , find the equation of the circles.

Let $A B$ be a chord of length 12 of the circle $(x - 2)^{2} + (y + 1)^{2} = \frac{169}{4}$ . If tangents drawn to the circle at points $A$ and $B$ intersect at the point $P$ , then five times the distance of point $P$ from chord $A B$ is equal to ________

Select Answer:

Visualized Solution

Identify Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​ and Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Equation of Tangent at a Point

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(2,1) into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=0

Simplify to get Tangent Equation

Introduce Circle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​ and the Chord

Perpendicular Distance Formula

Substitute Values for Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Calculate Perpendicular Distance .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Pythagoras Theorem in the Circle

Substitute into Pythagoras Theorem

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Tangent as a Bridge

The Geometry of the Chord

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The line 2x−y+1=0 is a tangent to the circle at the point (2,5) and the centre of the circle lies on x−2y=4. Then, the radius of the circle is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a circle C passing through the point (4,0) touches the circle x2+y2+4x−6y=12 externally at the point (1, -1), then the radius of C is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two circles, each of radius 5 units, touch each other at (1,2). If the equation of their common tangent is 4x+3y=10, find the equation of the circles.

Identify Circle $C_{1}$ and Point $P$

Substitute $P (2, 1)$ into $T = 0$

Introduce Circle $C_{2}$ and the Chord

Substitute Values for Distance $p$

Calculate Perpendicular Distance $p$

Calculate $R^{2}$

The line $2 x - y + 1 = 0$ is a tangent to the circle at the point $(2, 5)$ and the centre of the circle lies on $x - 2 y = 4$ . Then, the radius of the circle is:

If a circle C passing through the point (4,0) touches the circle $x^{2} + y^{2} + 4 x - 6 y = 12$ externally at the point (1, -1), then the radius of C is :

Two circles, each of radius 5 units, touch each other at $(1, 2)$ . If the equation of their common tangent is $4 x + 3 y = 10$ , find the equation of the circles.