JEE Main 2021 (31 August Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Circles: If the variable line $3 x + 4 y = α$ lies between the two circles $(x - 1)^{2} + (y - 1)^{2} = 1$ and $(x - 9)^{2} + (y - 1)^{2} = 4$ without intercepting a chord on either circle, then the sum of all the integral values of $α$ is .

Q: Why do we use the product of the evaluations $L(C_1) \cdot L(C_2) < 0$ to say the line is between the circles?

Think of the line $L(x, y) = 3x + 4y - \alpha = 0$ as a boundary. If you plug a point into the expression $3x + 4y - \alpha$, the sign tells you which side of the line the point lies on. If $L(C_1)$ is positive and $L(C_2)$ is negative (or vice versa), the line must pass between them. Their product being negative is the most elegant mathematical way to enforce this 'opposite sides' condition.

Q: Why is it $\ge$ and not $>$ for the distance condition? Doesn't the line have to be outside the circle?

Great catch! The problem states the line does not intercept a chord. A chord is a line segment connecting two points on a circle. If the line is tangent to the circle, it touches at exactly one point, which is not a chord. Therefore, the distance being exactly equal to the radius is perfectly acceptable. We only exclude the case where the distance is strictly less than the radius, which would create a secant line.

Q: How do I handle the intersection of these three intervals without getting confused?

The best way is to draw a number line. Plot the interval $(7, 31)$ from the first condition. Then, overlay the regions from the other two conditions. The region where all three conditions overlap is your final answer. It is a visual process, so don't rush it!

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: Visualize the line as a fence. For it to be 'between' the two circles, the centers of the circles must lie on opposite sides of the line. What does this imply about the sign of the expression

3 x + 4 y - α

when evaluated at the two centers?

Hint 2

Hint 2: The 'no chord' condition is the classic 'no intersection' condition. A line intersects a circle if the perpendicular distance from the center is less than the radius. Therefore, for the line to NOT intercept a chord, the perpendicular distance from the center to the line must be greater than or equal to the radius. Remember the formula:

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

Hint 3

Hint 3: Set up the three inequalities:

(3 (1) + 4 (1) - α) (3 (9) + 4 (1) - α) < 0

\frac{∣3 ( 1 ) + 4 ( 1 ) - α ∣}{5} \geq 1

, and

\frac{∣3 ( 9 ) + 4 ( 1 ) - α ∣}{5} \geq 2

. Solve these individually to find the valid range for

α

and then find their intersection.

Visualized Solution

The Geometric Setup

Two circles $C_{1}$ and $C_{2}$ are given.
A variable line $L : 3 x + 4 y = α$ passes between them.
Goal: Find the sum of all integral values of $α$ .

Properties of Circle $C_{1}$

Equation: $(x - 1)^{2} + (y - 1)^{2} = 1$
Center $C_{1} \equiv (1, 1)$
Radius $r_{1} = 1$

Properties of Circle $C_{2}$

Equation: $(x - 9)^{2} + (y - 1)^{2} = 4$
Center $C_{2} \equiv (9, 1)$
Radius $r_{2} = 2$

The Variable Line $L$

Line $L : 3 x + 4 y - α = 0$
The line must lie strictly between $C_{1}$ and $C_{2}$ .
This means $C_{1}$ and $C_{2}$ lie on opposite sides of $L$ .

Opposite Sides Condition

For opposite sides: $L (C_{1}) \cdot L (C_{2}) < 0$
$L (1, 1) = 3 (1) + 4 (1) - α = 7 - α$
$L (9, 1) = 3 (9) + 4 (1) - α = 31 - α$
$(7 - α) (31 - α) < 0$

Solving for $α$ (Condition 1)

$(7 - α) (31 - α) < 0$
$(α - 7) (α - 31) < 0$
$⟹ α \in (7, 31)$

No Chord on $C_{1}$

Line $L$ does not intercept a chord on $C_{1}$ .
Distance from center $C_{1}$ to line $L$ must be $\geq$ radius $r_{1}$ .
$d_{1} = \frac{∣3 ( 1 ) + 4 ( 1 ) - α ∣}{3 ^{2} + 4 ^{2}} \geq 1$

Solving for $α$ (Condition 2)

$\frac{∣7 - α ∣}{5} \geq 1 ⟹ ∣7 - α ∣ \geq 5$
Case 1: $7 - α \geq 5 ⟹ α \leq 2$
Case 2: $7 - α \leq - 5 ⟹ α \geq 12$
$α \in (- \infty, 2] \cup [12, \infty)$

No Chord on $C_{2}$

Line $L$ does not intercept a chord on $C_{2}$ .
Distance from center $C_{2}$ to line $L$ must be $\geq$ radius $r_{2}$ .
$d_{2} = \frac{∣3 ( 9 ) + 4 ( 1 ) - α ∣}{3 ^{2} + 4 ^{2}} \geq 2$

Solving for $α$ (Condition 3)

$\frac{∣31 - α ∣}{5} \geq 2 ⟹ ∣31 - α ∣ \geq 10$
Case 1: $31 - α \geq 10 ⟹ α \leq 21$
Case 2: $31 - α \leq - 10 ⟹ α \geq 41$
$α \in (- \infty, 21] \cup [41, \infty)$

Intersection of All Conditions

Condition 1: $α \in (7, 31)$
Condition 2: $α \in (- \infty, 2] \cup [12, \infty)$
Condition 3: $α \in (- \infty, 21] \cup [41, \infty)$
Intersection: $α \in [12, 21]$

Final Sum of Integral Values

Integral values: ${12, 13, 14, \dots, 21}$
This is an Arithmetic Progression (AP).
Number of terms $n = 21 - 12 + 1 = 10$
Sum $= \frac{n}{2} (a + l) = \frac{10}{2} (12 + 21)$
Sum $= 5 \times 33 = 165$

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We are tasked with finding the sum of all integral values of

α

such that the line

L : 3 x + 4 y - α = 0

separates the two circles

C_{1}

and

C_{2}

without intersecting them.

The first circle

C_{1}

is defined by

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

, which has center

C_{1} (1, 1)

and radius

r_{1} = 1

The second circle

C_{2}

is defined by

(x - 9)^{2} + (y - 1)^{2} = 4

, which has center

C_{2} (9, 1)

and radius

r_{2} = 2

The Condition of Separation

For the line to lie between the circles, the centers

C_{1}

and

C_{2}

must lie on opposite sides of the line

L

. This implies that the product of the line equation evaluated at these points must be negative:

L (1, 1) \cdot L (9, 1) < 0

Substituting the coordinates into the expression

3 x + 4 y - α

(3 (1) + 4 (1) - α) (3 (9) + 4 (1) - α) < 0

(7 - α) (31 - α) < 0

Solving this quadratic inequality, we find that

α

must lie in the open interval:

α \in (7, 31)

The No-Chord Constraint

To ensure the line does not intersect the circles, the perpendicular distance from each center to the line must be greater than or equal to the respective radius.

For

C_{1}

, the distance

d_{1}

is given by:

d_{1} = \frac{∣3 ( 1 ) + 4 ( 1 ) - α ∣}{3 ^{2} + 4 ^{2}} = \frac{∣7 - α ∣}{5}

Setting

d_{1} \geq 1

, we obtain

∣7 - α ∣ \geq 5

, which implies

α \leq 2

α \geq 12

For

C_{2}

, the distance

d_{2}

is given by:

d_{2} = \frac{∣3 ( 9 ) + 4 ( 1 ) - α ∣}{5} = \frac{∣31 - α ∣}{5}

Setting

d_{2} \geq 2

, we obtain

∣31 - α ∣ \geq 10

, which implies

α \leq 21

α \geq 41

The Grand Intersection

We must satisfy all three conditions simultaneously: 1.

α \in (7, 31)

α \in (- \infty, 2] \cup [12, \infty)

α \in (- \infty, 21] \cup [41, \infty)

Intersecting these sets, we observe that the valid range for

α

is:

α \in [12, 21]

Final Calculation

The integral values of

α

are

{12, 13, 14, 15, 16, 17, 18, 19, 20, 21}

. This is an arithmetic progression with

10

terms.

The sum of these values is calculated as:

S = \frac{10}{2} (12 + 21) = 5 \times 33 = 165

The final sum of all possible integral values of

α

is 165.

Common Traps & FAQs

Why do we use the product of the evaluations

L (C_{1}) \cdot L (C_{2}) < 0

to say the line is between the circles?

Think of the line

L (x, y) = 3 x + 4 y - α = 0

as a boundary. If you plug a point into the expression

3 x + 4 y - α

, the sign tells you which side of the line the point lies on. If

L (C_{1})

is positive and

L (C_{2})

is negative (or vice versa), the line must pass between them. Their product being negative is the most elegant mathematical way to enforce this 'opposite sides' condition.

Why is it

\geq

and not

>

for the distance condition? Doesn't the line have to be outside the circle?

Great catch! The problem states the line does not intercept a chord. A chord is a line segment connecting two points on a circle. If the line is tangent to the circle, it touches at exactly one point, which is not a chord. Therefore, the distance being exactly equal to the radius is perfectly acceptable. We only exclude the case where the distance is strictly less than the radius, which would create a secant line.

How do I handle the intersection of these three intervals without getting confused?

The best way is to draw a number line. Plot the interval

(7, 31)

from the first condition. Then, overlay the regions from the other two conditions. The region where all three conditions overlap is your final answer. It is a visual process, so don't rush it!

Animated Solution for Mathematics - Circles: If the variable line $3 x + 4 y = α$ lies between the two circles $(x - 1)^{2} + (y - 1)^{2} = 1$ and $(x - 9)^{2} + (y - 1)^{2} = 4$ without intercepting a chord on either circle, then the sum of all the integral values of $α$ is .

Enter Numerical Value:

Visualized Solution

The Geometric Setup

Properties of Circle $C_{1}$

Properties of Circle $C_{2}$

The Variable Line $L$

Opposite Sides Condition

Solving for $α$ (Condition 1)

No Chord on $C_{1}$

Solving for $α$ (Condition 2)

No Chord on $C_{2}$

Solving for $α$ (Condition 3)

Intersection of All Conditions

Final Sum of Integral Values

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Condition of Separation

The No-Chord Constraint

The Grand Intersection

Final Calculation

Similar Questions

The number of integral values of $k$ for which the line, $3 x + 4 y = k$ intersects the circle, $x^{2} + y^{2} - 2 x - 4 y + 4 = 0$ at two distinct points is

Let the centre of a circle $C$ be $(α, β)$ and its radius $r < 8$ . Let $3 x + 4 y = 24$ and $3 x - 4 y = 32$ be two tangents and $4 x + 3 y = 1$ be a normal to $C$ . Then $(α - β + r)$ is equal to

Equation of two diameters of a circle are $2 x - 3 y = 5$ and $3 x - 4 y = 7$ . The line joining the points $(- \frac{22}{7}, - 4)$ and $(- \frac{1}{7}, 3)$ intersects the circle at only one point $P (α, β)$ . Then $17 β - α$ is equal to

Two circles each of radius 5 units touch each other at the point $(1, 2)$ . If the equation of their common tangent is $4 x + 3 y = 10$ , and $C_{1} (α, β)$ and $C_{2} (γ, δ)$ , $C_{1} \neq = C_{2}$ are their centres, then $∣ (α + β) (γ + δ) ∣$ is equal to .

The circle $x^{2} + y^{2} = 4 x + 8 y + 5$ intersects the line $3 x - 4 y = m$ at two distinct points if

Let a line perpendicular to the line $2 x - y = 10$ touch the parabola $y^{2} = 4 (x - 9)$ at the point $P$ . The distance of the point $P$ from the centre of the circle $x^{2} + y^{2} - 14 x - 8 y + 56 = 0$ is __________

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

The Geometric Setup

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

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

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

Opposite Sides Condition

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

No Chord on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C1​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α (Condition 2)

No Chord on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α (Condition 3)

Intersection of All Conditions

Final Sum of Integral Values

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The Condition of Separation

The No-Chord Constraint

The Grand Intersection

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The number of integral values of k for which the line, 3x+4y=k intersects the circle, x2+y2−2x−4y+4=0 at two distinct points is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the centre of a circle C be (α,β) and its radius r<8. Let 3x+4y=24 and 3x−4y=32 be two tangents and 4x+3y=1 be a normal to C. Then (α−β+r) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The circle x2+y2=4x+8y+5 intersects the line 3x−4y=m at two distinct points if

Properties of Circle $C_{1}$

Properties of Circle $C_{2}$

The Variable Line $L$

Solving for $α$ (Condition 1)

No Chord on $C_{1}$

Solving for $α$ (Condition 2)

No Chord on $C_{2}$

Solving for $α$ (Condition 3)

The number of integral values of $k$ for which the line, $3 x + 4 y = k$ intersects the circle, $x^{2} + y^{2} - 2 x - 4 y + 4 = 0$ at two distinct points is

Let the centre of a circle $C$ be $(α, β)$ and its radius $r < 8$ . Let $3 x + 4 y = 24$ and $3 x - 4 y = 32$ be two tangents and $4 x + 3 y = 1$ be a normal to $C$ . Then $(α - β + r)$ is equal to

The circle $x^{2} + y^{2} = 4 x + 8 y + 5$ intersects the line $3 x - 4 y = m$ at two distinct points if