JEE Main 2024 (06 Apr Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Circles: Let $C$ be the circle of minimum area touching the parabola $y = 6 - x^{2}$ and the lines $y = 3 ∣ x ∣$ . Then, which one of the following points lies on the circle $C$ ?

Q: Why do we need to check the vertex separately? Doesn't the discriminant cover all tangency cases?

The discriminant method ($D=0$) works perfectly for 'smooth' tangency where the curves share a common tangent line. However, at the vertex of a parabola, the curvature changes abruptly relative to the axis. Sometimes, the circle 'kisses' the vertex, and the algebraic discriminant might not capture this specific geometric configuration. Always verify the vertex case as a separate, critical boundary condition!

Select Answer:

Visualized Solution

Visualizing the Setup

Parabola: $y = 6 - x^{2}$ (Vertex: $(0, 6)$ , opens downwards)
Lines: $y = 3 ∣ x ∣ ⟹ y = 3 x$ and $y = - 3 x$

Symmetry and Center

Due to symmetry about the $y$ -axis, the center of the circle $C$ must be $(0, k)$ .
Let the radius of the circle be $r$ .

Tangency to the Lines

Distance from $(0, k)$ to $3 x - y = 0$ is $r$ .
$r = \frac{∣ 3 ( 0 ) - k ∣}{( 3 ) ^{2} + ( - 1 ) ^{2}} = \frac{∣ - k ∣}{3 + 1} = \frac{k}{2}$
$⟹ k = 2 r$

Equation of the Circle

Circle Equation: $x^{2} + (y - 2 r)^{2} = r^{2}$
Substitute $x^{2} = 6 - y$ from the parabola equation into the circle equation:
$(6 - y) + (y - 2 r)^{2} = r^{2}$

Condition for Tangency

Expand: $6 - y + y^{2} - 4 r y + 4 r^{2} = r^{2}$
Rearrange as a quadratic in $y$ :
$y^{2} - (1 + 4 r) y + (6 + 3 r^{2}) = 0$
For tangency, the discriminant $D = 0$ .

Solving for the Radius

Discriminant: $(1 + 4 r)^{2} - 4 (1) (6 + 3 r^{2}) = 0$
$1 + 16 r^{2} + 8 r - 24 - 12 r^{2} = 0 ⟹ 4 r^{2} + 8 r - 23 = 0$
Solving for $r$ gives $r = - 1 \pm \frac{3 3}{2}$
Positive root $r \approx 1.598$ .

The Vertex Tangency Case

Tangency can also occur at the vertex $(0, 6)$ without $D = 0$ .
The top of the circle is at $y = 2 r + r = 3 r$ .
Equating to vertex height: $3 r = 6 ⟹ r = 2$ .

Checking the Vertex Case

If $r = 2$ , Center is $(0, 4)$ . Circle: $x^{2} + (y - 4)^{2} = 4$ .
Intersection with $y = 6 - x^{2}$ :
$(6 - y) + (y - 4)^{2} = 4 ⟹ y^{2} - 9 y + 18 = 0$
$(y - 3) (y - 6) = 0 ⟹ y = 3, 6$ .

Final Verification

Circle $C$ : $x^{2} + (y - 4)^{2} = 4$
Check point $(2, 4)$ : $2^{2} + (4 - 4)^{2} = 4 + 0 = 4$
The point $(2, 4)$ satisfies the circle equation.
Correct Option: (2, 4)

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The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

The problem involves a parabola defined by

y = 6 - x^{2}

and two lines defined by

y = 3 ∣ x ∣

. We seek a circle that is tangent to both the parabola and these lines.

The Symmetry Anchor

The first observation is the inherent symmetry of the system. Both the parabola and the V-shaped lines are symmetric about the

y

-axis.

Consequently, the center of the circle must lie on the

y

-axis. Let the center be

(0, k)

and the radius be

r

The Distance Constraint

The lines are given by

y = 3 x

and

y = - 3 x

, which can be rewritten as

3 x - y = 0

and

3 x + y = 0

. For the circle to be tangent to these lines, the perpendicular distance from

(0, k)

to the line

3 x - y = 0

must equal

r

Using the point-to-line distance formula:

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

This yields the relationship

k = 2 r

. Thus, the center of the circle is

(0, 2 r)

The Algebraic Dance

The equation of the circle is

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

. Given the parabola

x^{2} = 6 - y

, we substitute

x^{2}

into the circle's equation:

(6 - y) + (y - 2 r)^{2} = r^{2}

Expanding this expression, we obtain:

6 - y + y^{2} - 4 r y + 4 r^{2} = r^{2}

y^{2} - (1 + 4 r) y + (6 + 3 r^{2}) = 0

For the circle to be tangent to the parabola, the discriminant

D

of this quadratic must be zero:

D = (1 + 4 r)^{2} - 4 (1) (6 + 3 r^{2}) = 0

1 + 8 r + 16 r^{2} - 24 - 12 r^{2} = 0

4 r^{2} + 8 r - 23 = 0

The Vertex Trap

Tangency can also occur at the vertex

(0, 6)

of the parabola. The highest point of the circle is at

y = 2 r + r = 3 r

If the circle touches the vertex, then

3 r = 6

, which implies

r = 2

. For

r = 2

, the center is

(0, 4)

and the circle equation is

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

Testing this circle, we find it is tangent to the parabola at the vertex. This confirms that $r = 2$ is a valid solution, resulting in a circle centered at

(0, 4)

with radius

2

Common Traps & FAQs

Why can we assume the center is at

(0, k)

? Couldn't it be off-center?

Great question! Because the entire system—the parabola and the two lines—is perfectly symmetric about the

y

-axis, any circle touching both must also respect this symmetry. If the center were at

(h, k)

with

h \neq = 0

, you would need a mirror-image circle at

(- h, k)

to maintain the symmetry of the setup. Since we are looking for a single circle, its center must lie on the axis of symmetry.

Why do we need to check the vertex separately? Doesn't the discriminant cover all tangency cases?

The discriminant method (

D = 0

) works perfectly for 'smooth' tangency where the curves share a common tangent line. However, at the vertex of a parabola, the curvature changes abruptly relative to the axis. Sometimes, the circle 'kisses' the vertex, and the algebraic discriminant might not capture this specific geometric configuration. Always verify the vertex case as a separate, critical boundary condition!

How do I know which circle is the 'minimum area' one?

The problem asks for the circle of minimum area. After finding the possible radii, calculate the area for each valid case. The one with the smaller radius will naturally yield the smaller area. In this problem, we found a radius from the discriminant and a radius from the vertex condition; comparing them allows us to select the correct one.

Animated Solution for Mathematics - Circles: Let $C$ be the circle of minimum area touching the parabola $y = 6 - x^{2}$ and the lines $y = 3 ∣ x ∣$ . Then, which one of the following points lies on the circle $C$ ?

Select Answer:

Visualized Solution

Visualizing the Setup

Symmetry and Center

Tangency to the Lines

Equation of the Circle

Condition for Tangency

Solving for the Radius

The Vertex Tangency Case

Checking the Vertex Case

Final Verification

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Symmetry Anchor

The Distance Constraint

The Algebraic Dance

The Vertex Trap

Similar Questions

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Let the circle $C_{1} : x^{2} + y^{2} - 2 (x + y) + 1 = 0$ and $C_{2}$ be a circle having centre at $(- 1, 0)$ and radius 2. If the line of the common chord of $C_{1}$ and $C_{2}$ intersects the $y$ -axis at the point $P$ , then the square of the distance of $P$ from the centre of $C_{1}$ is :

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The circle passing through $(1, - 2)$ and touching the axis of $x$ at $(3, 0)$ also passes through the point

The points of intersection of the line $4 x - 3 y - 10 = 0$ and the circle $x^{2} + y^{2} - 2 x + 4 y - 20 = 0$ are ......... and .........

Let $C_{1}$ and $C_{2}$ be the centres of the circles $x^{2} + y^{2} - 2 x - 2 y - 2 = 0$ and $x^{2} + y^{2} - 6 x - 6 y + 14 = 0$ respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral $P C_{1} Q C_{2}$ is :

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A rectangle $R$ with end points of the one of its sides as $(1, 2)$ and $(3, 6)$ is inscribed in a circle. If the equation of a diameter of the circle is $2 x - y + 4 = 0$ , then the area of $R$ is _______.

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Select Answer:

Visualized Solution

Visualizing the Setup

Symmetry and Center

Tangency to the Lines

Equation of the Circle

Condition for Tangency

Solving for the Radius

The Vertex Tangency Case

Checking the Vertex Case

Final Verification

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Symmetry Anchor

The Distance Constraint

The Algebraic Dance

The Vertex Trap

Similar Questions

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