JEE Main 2020 - 8 Jan (Evening)

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

Select Answer:

Progressive Hints

Hint 1

Visualize the geometry. You have two distinct circles. The first circle is centered at the origin, and the second is shifted. The key isn't just the algebra; it's realizing that the line

L_{1}

acts as a geometric bridge between the two circles. Focus on the relationship between the tangent and the radius.

Hint 2

Recall your toolkit: The equation of a tangent to

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

(x_{1}, y_{1})

x x_{1} + y y_{1} = a^{2}

. Once you find the slope of

L_{1}

, use the condition

m_{1} \cdot m_{2} = - 1

to find the slope of your target line. This is the pivot point of the entire problem.

Hint 3

Now, for the final act: The line

L

is tangent to the second circle

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

. Use the perpendicular distance formula from the center

(3, 0)

to the line

x - y + c = 0

and set it equal to the radius,

1

. This will yield the quadratic equation you need.

Visualized Solution

Visualizing the Geometry

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

Tangent $L_{1}$ at $(\frac{1}{2}, \frac{1}{2})$

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

Equation of Tangent $L_{1}$

Formula for tangent to $x^{2} + y^{2} = a^{2}$ at $(x_{1}, y_{1})$ is $x x_{1} + y y_{1} = a^{2}$ .
Substitute $(x_{1}, y_{1}) = (\frac{1}{2}, \frac{1}{2})$ and $a^{2} = 1$ .
$\frac{x}{2} + \frac{y}{2} = 1$

Slope of $L_{1}$

Multiply by $2$ : $x + y = 2$
Rearrange to $y = m x + c$ form: $y = - x + 2$
Slope of $L_{1}$ ( $m_{1}$ ) $= - 1$

Finding Slope $m$ of Line $L$

Line $L$ is perpendicular to $L_{1}$ .
Condition for perpendicular lines: $m \cdot m_{1} = - 1$
Substitute $m_{1} = - 1$ : $m \cdot (- 1) = - 1 ⟹ m = 1$

Equation of Line $L$

Substitute $m = 1$ into $y = m x + c$ :
Line $L$ : $y = x + c$
General form: $x - y + c = 0$

Tangency Condition for Circle 2

Circle 2: $(x - 3)^{2} + y^{2} = 1$
Center $(h, k) = (3, 0)$ , Radius $r = 1$
Condition: Perpendicular distance from center $(3, 0)$ to line $L$ must equal the radius $1$ .

Applying the Distance Formula

Distance formula: $d = \frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}$
Substitute $(3, 0)$ into $x - y + c = 0$ :
$\frac{∣3 - 0 + c ∣}{1 ^{2} + ( - 1 ) ^{2}} = 1$

Simplifying the Equation

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

Squaring Both Sides

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

Expanding the Quadratic

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

Final Result

Subtract $2$ from both sides to form a standard quadratic equation:
$c^{2} + 6 c + 9 - 2 = 0$
$c^{2} + 6 c + 7 = 0$
Correct Option: (0)

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We are working with two circles on the coordinate plane. The first circle is centered at

(0, 0)

with radius

1

, defined by the equation

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

. The second circle is centered at

(3, 0)

with radius

1

, defined by

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

Phase 1

The Grazer and the Slope

We consider a point

P

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

on the first circle. The equation of the tangent line

L_{1}

at point

(x_{1}, y_{1})

for a circle

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

is given by

x x_{1} + y y_{1} = a^{2}

Substituting the coordinates of

P

, we obtain:

\frac{x}{2} + \frac{y}{2} = 1

Multiplying by

2

, we simplify this to

x + y = 2

, which can be rewritten as

y = - x + 2

. Thus, the slope of the tangent line

L_{1}

m_{1} = - 1

Phase 2

The Perpendicular Pivot

We seek a line

L

that is perpendicular to

L_{1}

. If two lines are perpendicular, the product of their slopes must be

- 1

Given

m_{1} = - 1

, the slope

m

of our target line

L

must satisfy:

m \cdot (- 1) = - 1 \Rightarrow m = 1

We can express the equation of line

L

in the slope-intercept form as

y = x + c

, or in the general form as:

x - y + c = 0

Phase 3

The Tangency Climax

For the line

x - y + c = 0

to be tangent to the second circle

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

, the perpendicular distance from the center

(3, 0)

to the line must equal the radius

r = 1

Using the distance formula

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

, we substitute the center

(3, 0)

and the line coefficients:

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

This simplifies to:

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

Final Calculation

To solve for

c

, we square both sides of the equation

∣ c + 3∣ = 2

(c + 3)^{2} = 2

Expanding the left side, we get:

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

Subtracting

2

from both sides, we arrive at the final quadratic equation:

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

Common Traps & FAQs

Wait, why did we use the point

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

for the first circle but not the second?

That is a fantastic observation! The problem explicitly defines

L_{1}

as the tangent to the *first* circle at that specific point. We use that point to define the slope of

L_{1}

, which then dictates the slope of our target line

L

via the perpendicularity condition. The second circle is just the destination where we apply that slope. Don't mix your circles!

Why is the perpendicular distance from the center to the line equal to the radius?

Think of the radius as the shortest path from the center to the tangent line. By definition, a tangent touches the circle at exactly one point, and the radius at that point is perpendicular to the tangent. If the distance were anything else, the line would either miss the circle entirely or cut through it as a secant. It is the fundamental definition of tangency!

Why did we square both sides to solve

∣ c + 3∣ = 2

It is the most elegant way to eliminate the absolute value. Since both sides are non-negative, squaring preserves the equality and gives us a clean quadratic equation. It is a standard, powerful technique in coordinate geometry to handle absolute values without splitting into cases.

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

Select Answer:

Visualized Solution

Visualizing the Geometry

Tangent $L_{1}$ at $(\frac{1}{2}, \frac{1}{2})$

Equation of Tangent $L_{1}$

Slope of $L_{1}$

Finding Slope $m$ of Line $L$

Equation of Line $L$

Tangency Condition for Circle 2

Applying the Distance Formula

Simplifying the Equation

Squaring Both Sides

Expanding the Quadratic

Final Result

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Final Calculation

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 $(\frac{1}{2}, \frac{1}{2})$ ; then:

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 :

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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

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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

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:

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.

Select Answer:

Visualized Solution

Visualizing the Geometry

Tangent .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (2​1​,2​1​)

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

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

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

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

Tangency Condition for Circle 2

Applying the Distance Formula

Simplifying the Equation

Squaring Both Sides

Expanding the Quadratic

Final Result

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Final Calculation

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

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.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

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.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

.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:

Tangent $L_{1}$ at $(\frac{1}{2}, \frac{1}{2})$

Equation of Tangent $L_{1}$

Slope of $L_{1}$

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Equation of Line $L$

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 :

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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

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

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: