JEE Main 2022 (29 July Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - 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 ________

Enter Numerical Value:

Visualized Solution

Equation of the Circle

Given circle: $(x - 2)^{2} + (y + 1)^{2} = \frac{169}{4}$
Standard form: $(x - h)^{2} + (y - k)^{2} = R^{2}$
Center $C = (2, - 1)$
Radius $R = \frac{169}{4} = \frac{13}{2}$

The Chord $A B$

A chord $A B$ of length $12$ is drawn.

Perpendicular from Center

Let $M$ be the midpoint of $A B$ .
The line joining the center to the midpoint is perpendicular to the chord: $C M ⊥ A B$ .
Therefore, $A M = \frac{12}{2} = 6$ .

Right Triangle $△ A M C$

Connect center $C$ to point $A$ to form $△ A M C$ .
$△ A M C$ is a right-angled triangle at $M$ .
Hypotenuse $A C = R = \frac{13}{2}$ .

Calculating Distance $C M$

Using Pythagoras theorem in $△ A M C$ :
$C M^{2} + A M^{2} = A C^{2}$
$C M = A C^{2} - A M^{2}$

Computing $C M$

Substitute the values:
$C M = (\frac{13}{2})^{2} - 6^{2}$
$C M = \frac{169}{4} - 36 = \frac{169 - 144}{4}$
$C M = \frac{25}{4} = \frac{5}{2}$

Tangents Intersecting at $P$

Tangents drawn at points $A$ and $B$ intersect at point $P$ .
By symmetry, $P$ lies on the extended line $C M$ .

Right Triangle $△ P A C$

The radius is perpendicular to the tangent at the point of contact.
Therefore, $∠ P A C = 9 0^{\circ}$ .
In right $△ P A C$ , $A M$ is the altitude to the hypotenuse $P C$ .

Altitude to Hypotenuse Property

For a right triangle with an altitude drawn to the hypotenuse:
$A M^{2} = P M \cdot C M$

Solving for Distance $P M$

Substitute $A M = 6$ and $C M = \frac{5}{2}$ :
$6^{2} = P M \cdot \frac{5}{2}$
$36 = P M \cdot \frac{5}{2}$
$P M = \frac{72}{5}$

Final Answer

The question asks for $5$ times the distance of $P$ from chord $A B$ .
Distance $= P M = \frac{72}{5}$
Required value $= 5 \times P M = 5 \times \frac{72}{5} = 72$

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We are given the circle equation

(x - 2)^{2} + (y + 1)^{2} = \frac{169}{4}

. This is in the standard form

(x - h)^{2} + (y - k)^{2} = R^{2}

By inspection, the center

C

(2, - 1)

and the radius

R

is:

R = \frac{169}{4} = \frac{13}{2} = 6.5

We introduce a chord

A B

with a length of

12

. We drop a perpendicular from the center

C

to the chord

A B

, meeting it at point

M

Since the perpendicular from the center to a chord bisects the chord, the segment

A M

is:

A M = \frac{12}{2} = 6

The Perpendicular Distance

Consider the right-angled triangle

△ A M C

, where

A C

is the hypotenuse (the radius). Using the Pythagorean theorem, we find the distance

C M

from the center to the chord:

C M = A C^{2} - A M^{2}

C M = (6.5)^{2} - 6^{2} = 42.25 - 36 = 6.25 = 2.5

Thus, the distance

C M = \frac{5}{2}

The Tangents and the Point $P$

Tangents drawn at points

A

and

B

meet at an external point

P

. Due to symmetry,

P

lies on the line extending from

C

through

M

Consider the right-angled triangle

△ P A C

, where

∠ P A C = 9 0^{\circ}

because the radius

A C

is perpendicular to the tangent

P A

. In this triangle,

A M

is the altitude dropped from the right angle

A

to the hypotenuse

P C

The Master Equation

The altitude to the hypotenuse creates similar triangles, specifically

△ P M A \sim △ A M C

. This similarity yields the geometric relationship:

A M^{2} = P M \cdot C M

Substituting the known values

A M = 6

and

C M = 2.5

6^{2} = P M \cdot 2.5

36 = P M \cdot 2.5

P M = \frac{36}{2.5} = 14.4 = \frac{72}{5}

Final Calculation

The problem requires us to find five times the distance

P M

. Calculating this, we get:

5 \times P M = 5 \times \frac{72}{5} = 72

The final result is 72.

Common Traps & FAQs

Wait, why does point

P

lie on the line

C M

Great observation! Because the tangents from an external point

P

to a circle are equal in length (

P A = P B

), the triangle

△ P A B

is isosceles. The line from the center

C

P

is the angle bisector of

∠ A P B

, and by symmetry, it must also be the perpendicular bisector of the chord

A B

. Thus,

P

C

, and

M

are collinear.

Why is

△ P A C

a right-angled triangle?

This is a fundamental property of circles: the radius drawn to the point of tangency is always perpendicular to the tangent line. Since

A C

is the radius and

P A

is the tangent at

A

, the angle

∠ P A C

must be

9 0^{\circ}

How did you get

A M^{2} = P M \cdot C M

This comes from the similarity of triangles. In the right-angled triangle

△ P A C

A M

is the altitude to the hypotenuse

P C

. This creates two smaller triangles,

△ P M A

and

△ A M C

, which are both similar to the larger triangle

△ P A C

and to each other. From

△ P M A \sim △ A M C

, we get the ratio

\frac{P M}{A M} = \frac{A M}{C M}

, which simplifies to

A M^{2} = P M \cdot C M

Enter Numerical Value:

Visualized Solution

Equation of the Circle

The Chord $A B$

Perpendicular from Center

Right Triangle $△ A M C$

Calculating Distance $C M$

Computing $C M$

Tangents Intersecting at $P$

Right Triangle $△ P A C$

Altitude to Hypotenuse Property

Solving for Distance $P M$

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Perpendicular Distance

The Tangents and the Point $P$

The Master Equation

Final Calculation

Similar Questions

Circles with radii 3, 4 and 5 touch each other externally. If $P$ is the point of intersection of tangents to these circles at their points of contact, find the distance of $P$ from the points of contact.

If a line along a chord of the circle $4 x^{2} + 4 y^{2} + 120 x + 675 = 0$ , passes through the point $(- 30, 0)$ and is tangent to the parabola $y^{2} = 30 x$ , then the length of this chord is :

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 :-

Let a tangent to the Curve $9 x^{2} + 16 y^{2} = 144$ intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is _____.

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Let a circle $C$ of radius 1 and closer to the origin be such that the lines passing through the point $(3, 2)$ and parallel to the coordinate axes touch it. Then the shortest distance of the circle $C$ from the point $(5, 5)$ is :

Let the tangent to the parabola $y^{2} = 12 x$ at the point $(3, α)$ be perpendicular to the line $2 x + 2 y = 3$ . Then the square of distance of the point $(6, - 4)$ from the normal to the hyperbola $α^{2} x^{2} - 9 y^{2} = 9 α^{2}$ at its point $(α - 1, α + 2)$ is equal to .............

Enter Numerical Value:

Visualized Solution

Equation of the Circle

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

Perpendicular from Center

Right Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △AMC

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

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

Tangents Intersecting at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Right Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } △PAC

Altitude to Hypotenuse Property

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

Final Answer

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Perpendicular Distance

The Tangents and the Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

The Master Equation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a line along a chord of the circle 4x2+4y2+120x+675=0, passes through the point (−30,0) and is tangent to the parabola y2=30x, then the length of this chord is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The tangent to the circle C1​:x2+y2−2x−1=0 at the point (2, 1) cuts off a chord of length 4 from a circle C2​ whose centre is (3, -2). The radius of C2​ is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a tangent to the Curve 9x2+16y2=144 intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is _____.

The Chord $A B$

Right Triangle $△ A M C$

Calculating Distance $C M$

Computing $C M$

Tangents Intersecting at $P$

Right Triangle $△ P A C$

Solving for Distance $P M$

The Tangents and the Point $P$

If a line along a chord of the circle $4 x^{2} + 4 y^{2} + 120 x + 675 = 0$ , passes through the point $(- 30, 0)$ and is tangent to the parabola $y^{2} = 30 x$ , then the length of this chord is :

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 :-

Let a tangent to the Curve $9 x^{2} + 16 y^{2} = 144$ intersect the coordinate axes at the points A and B. Then, the minimum length of the line segment AB is _____.