JEE Main 2020 (8 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Circles: If a line, $y = m x + c$ is a tangent to the circle, $(x - 3)^{2} + y^{2} = 1$ and it is perpendicular to a line $L_{1}$ , where $L_{1}$ is the tangent to the circle, $x^{2} + y^{2} = 1$ at the point $(\frac{1}{2}, \frac{1}{2})$ ; then:

Q: Wait, why is the slope of $L_1$ equal to $-1$?

Great question! The tangent $L_1$ touches the circle $x^2 + y^2 = 1$ at the point $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$. The slope of the radius connecting the origin $(0, 0)$ to this point is $\frac{1/\sqrt{2} - 0}{1/\sqrt{2} - 0} = 1$. Since the tangent is always perpendicular to the radius at the point of contact, the slope of the tangent must be the negative reciprocal of the radius's slope, which is $-1$.

Q: Wait, why do we square both sides? Can't we just solve $|3 + c| = \sqrt{2}$?

You absolutely can! Solving $|3 + c| = \sqrt{2}$ gives two possible values: $3 + c = \sqrt{2}$ or $3 + c = -\sqrt{2}$. However, the question asks for a quadratic equation in $c$. Squaring both sides, $(3 + c)^2 = 2$, naturally leads to the quadratic form $c^2 + 6c + 9 = 2$, which simplifies to $c^2 + 6c + 7 = 0$. It is a direct path to the required answer format.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the two circles on the coordinate plane. The first circle is centered at the origin, and the second is shifted to the right. The key is to understand that the tangent line

L_{1}

is a geometric constraint on the first circle, and the line

L_{2}

is a geometric constraint on the second circle, linked by their perpendicularity.

Hint 2

Hint 2 (The Arsenal): Recall that for a line

y = m x + c

to be tangent to a circle

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

, the perpendicular distance from the center

(h, k)

to the line must be equal to the radius

r

. Also, remember that if two lines are perpendicular, the product of their slopes is

- 1

Hint 3

Hint 3 (The Execution): Start by finding the slope of

L_{1}

at the given point using differentiation or the property of the radius. Once you have the slope of

L_{1}

, determine the slope of

L_{2}

using the perpendicularity condition. Finally, set up the distance formula from the center of the second circle

(3, 0)

to the line

L_{2}

and equate it to the radius

1

to solve for

c

Visualized Solution

Geometry of the Circles

Circle $C_{1}$ : $x^{2} + y^{2} = 1$ , Center $(0, 0)$ , Radius $1$ .
Circle $C_{2}$ : $(x - 3)^{2} + y^{2} = 1$ , Center $(3, 0)$ , Radius $1$ .

Point $P$ and Tangent $L_{1}$

Point $P$ on $C_{1}$ : $(\frac{1}{2}, \frac{1}{2})$ .
Line $L_{1}$ is tangent to $C_{1}$ at $P$ .

Slope of Tangent $L_{1}$

Differentiating $x^{2} + y^{2} = 1$ with respect to $x$ .
$2 x + 2 y \frac{d y}{d x} = 0 ⟹ \frac{d y}{d x} = - \frac{x}{y}$ .

Calculating $m_{L 1}$

Substitute $P (\frac{1}{2}, \frac{1}{2})$ into $\frac{d y}{d x}$ .
$m_{L 1} = - \frac{\frac{1}{2}}{\frac{1}{2}} = - 1$ .

Introducing Line $L_{2}$

Given line $L_{2}$ : $y = m x + c$ .
$L_{2}$ is perpendicular to $L_{1}$ .
Condition for perpendicular lines: $m \cdot m_{L 1} = - 1$ .

Slope of Line $L_{2}$

Substitute $m_{L 1} = - 1$ .
$m \cdot (- 1) = - 1 ⟹ m = 1$ .

Equation of Line $L_{2}$

Substitute $m = 1$ into $y = m x + c$ .
$y = 1 \cdot x + c$ .
Rearranging into general form: $x - y + c = 0$ .

Condition for Tangency on $C_{2}$

Line $L_{2}$ is a tangent to circle $C_{2}$ .
For tangency, the perpendicular distance from the center of $C_{2}$ to $L_{2}$ must equal the radius of $C_{2}$ .
Center of $C_{2}$ : $(3, 0)$ , Radius: $1$ .

Applying the Distance Formula

Distance formula: $d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$ .
Substitute $(x_{1}, y_{1}) = (3, 0)$ and line $x - y + c = 0$ .
$d = \frac{∣1 ( 3 ) - 1 ( 0 ) + c ∣}{1 ^{2} + ( - 1 ) ^{2}} = 1$ .

Simplifying the Numerator and Denominator

Numerator: $∣1 (3) - 0 + c ∣ = ∣3 + c ∣$ .
Denominator: $1^{2} + (- 1)^{2} = 2$ .
Equation becomes: $\frac{∣3 + c ∣}{2} = 1$ .

Isolating the Modulus

Multiply both sides by $2$ .
$∣3 + c ∣ = 2$ .

Squaring Both Sides

To remove the absolute value, square both sides.
$(3 + c)^{2} = (2)^{2}$ .
$(3 + c)^{2} = 2$ .

Expanding the Square

Expand $(3 + c)^{2}$ using $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ .
$9 + 6 c + c^{2} = 2$ .

Final Quadratic Equation

Rearrange terms to form a standard quadratic equation.
$c^{2} + 6 c + 9 - 2 = 0$ .
$c^{2} + 6 c + 7 = 0$ .

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The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at two circles. The first,

C_{1}

, is perfectly centered at the origin

(0, 0)

with a radius of

1

The second,

C_{2}

, is a mirror image of the first, but shifted along the x-axis to be centered at

(3, 0)

, also with a radius of

1

. This is our playground.

The First Tangent: $L_{1}$

Our journey begins with the first circle,

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

. We are given a point

P

on this circle at

(\frac{1}{2}, \frac{1}{2})

A tangent line,

L_{1}

, grazes the circle at this exact point. To find the slope of this line, we use calculus by differentiating the equation

x^{2} + y^{2} = 1

with respect to

x

2 x + 2 y \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{x}{y}

Substituting our point

P

, the slope

m_{L 1}

becomes:

m_{L 1} = - \frac{1/ 2}{1/ 2} = - 1

The Perpendicular Dance: $L_{2}$

Now, we introduce the line

L_{2}

, defined by

y = m x + c

. We are told that

L_{2}

is perpendicular to

L_{1}

In coordinate geometry, the condition for two lines to be perpendicular is that the product of their slopes must be

- 1

. Since

m_{L 1} = - 1

, we have

m \cdot (- 1) = - 1

, which forces

m = 1

Our line

L_{2}

now takes the form

y = x + c

, or in its general form:

x - y + c = 0

The Tangency Condition

The problem states that

L_{2}

is a tangent to the second circle,

C_{2} : (x - 3)^{2} + y^{2} = 1

. For any line to be tangent to a circle, the perpendicular distance from the center of the circle to the line must be exactly equal to the radius.

The center of

C_{2}

(3, 0)

and its radius is

1

. Using the distance formula

d = \frac{∣ A x _{0} + B y _{0} + C ∣}{A ^{2} + B ^{2}}

, we substitute our values:

d = \frac{∣1 ( 3 ) - 1 ( 0 ) + c ∣}{1 ^{2} + ( - 1 ) ^{2}} = 1

This simplifies to:

\frac{∣3 + c ∣}{2} = 1 \Rightarrow ∣3 + c ∣ = 2

The Final Quadratic

To eliminate the modulus and find the quadratic equation, we square both sides:

(3 + c)^{2} = (2)^{2}

Expanding the left side gives us

9 + 6 c + c^{2} = 2

. Rearranging the terms to set the equation to zero, we subtract

2

from

9

to get

7

Thus, we arrive at our final, elegant result:

$c^{2} + 6 c + 7 = 0$

Common Traps & FAQs

Wait, why is the slope of

L_{1}

equal to

- 1

Great question! The tangent

L_{1}

touches the circle

x^{2} + y^{2} = 1

at the point

(\frac{1}{2}, \frac{1}{2})

. The slope of the radius connecting the origin

(0, 0)

to this point is

\frac{1/ 2 - 0}{1/ 2 - 0} = 1

. Since the tangent is always perpendicular to the radius at the point of contact, the slope of the tangent must be the negative reciprocal of the radius's slope, which is

- 1

Why do we use the perpendicular distance formula instead of solving the quadratic equation of intersection?

You could solve for the intersection, but that would involve substituting

y = x + c

into the equation of the second circle, resulting in a quadratic in

x

, and then setting the discriminant to zero. While valid, using the perpendicular distance formula

d = \frac{∣ A x _{0} + B y _{0} + C ∣}{A ^{2} + B ^{2}}

is much faster and less prone to algebraic errors. It is the standard, elegant way to handle tangency conditions in coordinate geometry.

Wait, why do we square both sides? Can't we just solve

∣3 + c ∣ = 2

You absolutely can! Solving

∣3 + c ∣ = 2

gives two possible values:

3 + c = 2

3 + c = - 2

. However, the question asks for a quadratic equation in

c

. Squaring both sides,

(3 + c)^{2} = 2

, naturally leads to the quadratic form

c^{2} + 6 c + 9 = 2

, which simplifies to

c^{2} + 6 c + 7 = 0

. It is a direct path to the required answer format.

Animated Solution for Mathematics - Circles: If a line, $y = m x + c$ is a tangent to the circle, $(x - 3)^{2} + y^{2} = 1$ and it is perpendicular to a line $L_{1}$ , where $L_{1}$ is the tangent to the circle, $x^{2} + y^{2} = 1$ at the point $(\frac{1}{2}, \frac{1}{2})$ ; then:

Select Answer:

Visualized Solution

Geometry of the Circles

Point $P$ and Tangent $L_{1}$

Slope of Tangent $L_{1}$

Calculating $m_{L 1}$

Introducing Line $L_{2}$

Slope of Line $L_{2}$

Equation of Line $L_{2}$

Condition for Tangency on $C_{2}$

Applying the Distance Formula

Simplifying the Numerator and Denominator

Isolating the Modulus

Squaring Both Sides

Expanding the Square

Final Quadratic Equation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The First Tangent: $L_{1}$

The Perpendicular Dance: $L_{2}$

The Tangency Condition

The Final Quadratic

Similar Questions

If a line, $y = m x + c$ is a tangent to the circle, $(x - 3)^{2} + y^{2} = 1$ and it is perpendicular to a line $L_{1}$ , where $L_{1}$ is the tangent to the circle, $x^{2} + y^{2} = 1$ at the point $(\frac{1}{2}, \frac{1}{2})$ :

Comprehension Passage

A possible equation of $L$ is

A common tangent of the two circles is

If the tangent at $(1, 7)$ to the curve $x^{2} = y - 6$ touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ then the value of $c$ is :

Tangent to the curve $y = x^{2} + 6$ at a point $(1, 7)$ touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ at a point $Q$ . Then the coordinates of $Q$ are

The angle between a pair of tangents drawn from a point $P$ to the circle $x^{2} + y^{2} + 4 x - 6 y + 9 sin^{2} α + 13 cos^{2} α = 0$ is $2 α$ . The equation of the locus of the point $P$ is

Let $P Q$ and $M N$ be two straight lines touching the circle $x^{2} + y^{2} - 4 x - 6 y - 3 = 0$ at the points $A$ and $B$ respectively. Let $O$ be the centre of the circle and $∠ A O B = π /3$ . Then the locus of the point of intersection of the lines $P Q$ and $M N$ 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 a common tangent to the circle, $x^{2} + y^{2} - 6 x = 0$ and the parabola, $y^{2} = 4 x$ , is:

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is

Select Answer:

Visualized Solution

Geometry of the Circles

Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and Tangent .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Slope of Tangent .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mL1​

Introducing Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Slope of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Equation of Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Condition for Tangency on .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Applying the Distance Formula

Simplifying the Numerator and Denominator

Isolating the Modulus

Squaring Both Sides

Expanding the Square

Final Quadratic Equation

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

The First Tangent: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

The Perpendicular Dance: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

The Tangency Condition

The Final Quadratic

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A possible equation of L is

A common tangent of the two circles is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the tangent at (1,7) to the curve x2=y−6 touches the circle x2+y2+16x+12y+c=0 then the value of c is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tangent to the curve y=x2+6 at a point (1,7) touches the circle x2+y2+16x+12y+c=0 at a point Q. Then the coordinates of Q are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The angle between a pair of tangents drawn from a point P to the circle x2+y2+4x−6y+9sin2α+13cos2α=0 is 2α. The equation of the locus of the point P 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; } Equation of a common tangent to the circle, x2+y2−6x=0 and the parabola, y2=4x, is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a tangent to the circle x2+y2=1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is

Point $P$ and Tangent $L_{1}$

Slope of Tangent $L_{1}$

Calculating $m_{L 1}$

Introducing Line $L_{2}$

Slope of Line $L_{2}$

Equation of Line $L_{2}$

Condition for Tangency on $C_{2}$

The First Tangent: $L_{1}$

The Perpendicular Dance: $L_{2}$

A possible equation of $L$ is

If the tangent at $(1, 7)$ to the curve $x^{2} = y - 6$ touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ then the value of $c$ is :

Tangent to the curve $y = x^{2} + 6$ at a point $(1, 7)$ touches the circle $x^{2} + y^{2} + 16 x + 12 y + c = 0$ at a point $Q$ . Then the coordinates of $Q$ are

The angle between a pair of tangents drawn from a point $P$ to the circle $x^{2} + y^{2} + 4 x - 6 y + 9 sin^{2} α + 13 cos^{2} α = 0$ is $2 α$ . The equation of the locus of the point $P$ 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 a common tangent to the circle, $x^{2} + y^{2} - 6 x = 0$ and the parabola, $y^{2} = 4 x$ , is:

If a tangent to the circle $x^{2} + y^{2} = 1$ intersects the coordinate axes at distinct points $P$ and $Q$ , then the locus of the mid-point of $P Q$ is