JEE Main 2020 - 4 Sep (Evening)

LEVELJEE Main

Animated Solution for Mathematics - Circles: Let $P Q$ be a diameter of the circle $x^{2} + y^{2} = 9$ . If $α$ and $β$ are the lengths of the perpendiculars from $P$ and $Q$ on the straight line, $x + y = 2$ respectively, then the maximum value of $α β$ is

Q: Is there a geometric way to solve this without all the trigonometry?

Yes! You could consider the line $x+y=2$ and the circle. The product of distances from the ends of a diameter to a line is related to the power of the point or the distance from the center. However, the trigonometric method is the most direct path to the maximum value in JEE exams.

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e C i r c l e an d L in e$

Circle: $x^{2} + y^{2} = 9$ (Center $(0, 0)$ , Radius $r = 3$ )
Line: $x + y = 2$

$T h eD iam e t er P Q$

$P Q$ is a diameter of the circle.
It passes through the origin $(0, 0)$ .

$P a r am e t r i c C oor d ina t eso f P an d Q$

Let $P = (3 cos θ, 3 sin θ)$
Since $Q$ is diametrically opposite, $Q = (- 3 cos θ, - 3 sin θ)$

$P er p e n d i c u l a r s α an d β$

$α$ : Perpendicular distance from $P$ to $x + y = 2$
$β$ : Perpendicular distance from $Q$ to $x + y = 2$

$D i s t an ce F or m u l a$

Distance from $(x_{1}, y_{1})$ to $a x + b y + c = 0$ is $d = \frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}$
Line equation in standard form: $x + y - 2 = 0$

$E x p r ess i o n f or α$

For $P (3 cos θ, 3 sin θ)$ and line $x + y - 2 = 0$ :
$α = \frac{∣3 c o s θ + 3 s i n θ - 2∣}{1 ^{2} + 1 ^{2}}$
$α = \frac{∣3 c o s θ + 3 s i n θ - 2∣}{2}$

$E x p r ess i o n f or β$

For $Q (- 3 cos θ, - 3 sin θ)$ and line $x + y - 2 = 0$ :
$β = \frac{∣ - 3 c o s θ - 3 s i n θ - 2∣}{2}$
$β = \frac{∣3 c o s θ + 3 s i n θ + 2∣}{2}$

$P r o d u c t o f α an d β$

$α β = \frac{∣3 c o s θ + 3 s i n θ - 2∣}{2} \cdot \frac{∣3 c o s θ + 3 s i n θ + 2∣}{2}$
Combine the denominators: $2 \cdot 2 = 2$
Combine the numerators inside a single absolute value.

$A ppl y in g D i f f er e n ceo f S q u a r es$

Let $A = 3 (cos θ + sin θ)$ and $B = 2$
Numerator is $∣ (A - B) (A + B) ∣ = ∣ A^{2} - B^{2} ∣$
$α β = \frac{∣ ( 3 ( c o s θ + s i n θ ) ) ^{2} - 2 ^{2} ∣}{2}$

$E x p an d in g t h e S q u a r e d T er m$

Expand: $(cos θ + sin θ)^{2} = cos^{2} θ + sin^{2} θ + 2 sin θ cos θ$
Use identity: $cos^{2} θ + sin^{2} θ = 1$
Use identity: $2 sin θ cos θ = sin 2 θ$
Result: $(cos θ + sin θ)^{2} = 1 + sin 2 θ$

$S im pl i f i e d E x p r ess i o n f or α β$

Substitute back into the product:
$α β = \frac{∣9 ( 1 + s i n 2 θ ) - 4∣}{2}$
$α β = \frac{∣9 + 9 s i n 2 θ - 4∣}{2}$
$α β = \frac{∣5 + 9 s i n 2 θ ∣}{2}$

$M a x imi z in g t h e P r o d u c t$

To maximize $α β = \frac{∣5 + 9 s i n 2 θ ∣}{2}$ , we need the maximum value of the numerator.
The sine function is bounded: $- 1 \leq sin 2 θ \leq 1$
Maximum value occurs when $sin 2 θ = 1$

$F ina l C a l c u l a t i o n$

Substitute $sin 2 θ = 1$ :
$(α β)_{ma x} = \frac{∣5 + 9 ( 1 ) ∣}{2}$
$(α β)_{ma x} = \frac{14}{2} = 7$
Final Answer: $7$

00:00 / 00:00

The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

The problem involves a circle defined by the equation

x^{2} + y^{2} = 9

, which is centered at the origin with a radius

r = 3

. We are considering a diameter

P Q

of this circle and a line defined by

x + y - 2 = 0

We define

α

and

β

as the perpendicular distances from points

P

and

Q

to the line, respectively. Our objective is to determine the maximum value of the product

α β

Phase 1

Parametric Elegance

To simplify the geometry, we represent the coordinates of point

P

using trigonometry. Given the radius is

3

, we set:

P = (3 cos θ, 3 sin θ)

Since

Q

is diametrically opposite to

P

, its coordinates are the reflection of

P

through the origin:

Q = (- 3 cos θ, - 3 sin θ)

Phase 2

The Bridge of Distance

The perpendicular distance

d

from a point

(x_{1}, y_{1})

to the line

a x + b y + c = 0

is given by:

d = \frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}

Applying this to point

P

for the line

x + y - 2 = 0

α = \frac{∣3 cos θ + 3 sin θ - 2∣}{1 ^{2} + 1 ^{2}} = \frac{∣3 ( cos θ + sin θ ) - 2∣}{2}

Similarly, for point

Q

β = \frac{∣ - 3 cos θ - 3 sin θ - 2∣}{2} = \frac{∣3 ( cos θ + sin θ ) + 2∣}{2}

Phase 3

The Algebraic Triumph

We now calculate the product

α β

α β = \frac{∣3 ( cos θ + sin θ ) - 2∣ \cdot ∣3 ( cos θ + sin θ ) + 2∣}{2}

Using the difference of squares identity

(A - B) (A + B) = A^{2} - B^{2}

, where

A = 3 (cos θ + sin θ)

and

B = 2

, we obtain:

α β = \frac{∣9 ( cos θ + sin θ ) ^{2} - 4∣}{2}

Phase 4

The Final Flourish

We expand the term

(cos θ + sin θ)^{2}

using the identity

cos^{2} θ + sin^{2} θ + 2 sin θ cos θ = 1 + sin 2 θ

. Substituting this into our expression:

α β = \frac{∣9 ( 1 + sin 2 θ ) - 4∣}{2} = \frac{∣5 + 9 sin 2 θ ∣}{2}

To maximize the product, we set

sin 2 θ

to its maximum value of

1

. The calculation follows:

Max (α β) = \frac{∣5 + 9 ( 1 ) ∣}{2} = \frac{14}{2} = 7

The maximum value of the product

α β

is 7.

Common Traps & FAQs

Why can't we just use the distance from the center to the line and relate it to

α

and

β

That is a brilliant geometric intuition! The distance from the center

(0, 0)

to the line

x + y - 2 = 0

d = \frac{∣ - 2∣}{2} = 2

. While you can relate

α

and

β

to this distance using properties of the diameter, the algebraic approach using parametric coordinates is more robust for finding the *maximum* value without getting lost in geometric constraints.

Wait, why do we use absolute values? Can't we just drop them?

Great question! The absolute value is essential because distance must always be positive. If you drop the absolute value, you might end up with a negative product, which is physically impossible for lengths. Always keep the absolute value until you are certain of the sign of the expression inside.

Is there a geometric way to solve this without all the trigonometry?

Yes! You could consider the line

x + y = 2

and the circle. The product of distances from the ends of a diameter to a line is related to the power of the point or the distance from the center. However, the trigonometric method is the most direct path to the maximum value in JEE exams.

Animated Solution for Mathematics - Circles: Let $P Q$ be a diameter of the circle $x^{2} + y^{2} = 9$ . If $α$ and $β$ are the lengths of the perpendiculars from $P$ and $Q$ on the straight line, $x + y = 2$ respectively, then the maximum value of $α β$ is

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e C i r c l e an d L in e$

$T h eD iam e t er P Q$

$P a r am e t r i c C oor d ina t eso f P an d Q$

$P er p e n d i c u l a r s α an d β$

$D i s t an ce F or m u l a$

$E x p r ess i o n f or α$

$E x p r ess i o n f or β$

$P r o d u c t o f α an d β$

$A ppl y in g D i f f er e n ceo f S q u a r es$

$E x p an d in g t h e S q u a r e d T er m$

$S im pl i f i e d E x p r ess i o n f or α β$

$M a x imi z in g t h e P r o d u c t$

$F ina l C a l c u l a t i o n$

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

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Enter Numerical Value:

Visualized Solution

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

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

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

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

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

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

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

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

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

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

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

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

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

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The diameter of the circle, whose centre lies on the line x+y=2 in the first quadrant and which touches both the lines x=3 and y=2, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the locus of the mid points of the chords of circle x2+(y−1)2=1 drawn from the origin intersect the line x+y=1 at P and Q. Then, the length of PQ is :

$V i s u a l i z in g t h e C i r c l e an d L in e$

$T h eD iam e t er P Q$

$P a r am e t r i c C oor d ina t eso f P an d Q$

$P er p e n d i c u l a r s α an d β$

$D i s t an ce F or m u l a$

$E x p r ess i o n f or α$

$E x p r ess i o n f or β$

$P r o d u c t o f α an d β$

$A ppl y in g D i f f er e n ceo f S q u a r es$

$E x p an d in g t h e S q u a r e d T er m$

$S im pl i f i e d E x p r ess i o n f or α β$

$M a x imi z in g t h e P r o d u c t$

$F ina l C a l c u l a t i o n$

The diameter of the circle, whose centre lies on the line $x + y = 2$ in the first quadrant and which touches both the lines $x = 3$ and $y = 2$ , is

Let the locus of the mid points of the chords of circle $x^{2} + (y - 1)^{2} = 1$ drawn from the origin intersect the line $x + y = 1$ at $P$ and $Q$ . Then, the length of $P Q$ is :