JEE Main 2024 (05 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Circles: Let a circle $C$ of radius 1 and closer to the origin be such that the lines passing through the point $(3, 2)$ and parallel to the coordinate axes touch it. Then the shortest distance of the circle $C$ from the point $(5, 5)$ is :

Q: Why are there four possible centers? I thought a circle touching two lines was unique.

That is a fantastic observation! Think of the lines $x = 3$ and $y = 2$ as a cross. A circle of radius $1$ can be placed in any of the four corners created by this cross while still touching both lines. Because the absolute value equations $|h - 3| = 1$ and $|k - 2| = 1$ allow for both positive and negative offsets, we mathematically generate four distinct candidates: $(4, 3), (2, 3), (4, 1),$ and $(2, 1)$.

Q: Why do we subtract the radius at the end?

Imagine drawing a straight line from the external point $P$ to the center of the circle $C$. This line segment represents the distance between the point and the center. However, the circle is a boundary, not a solid point. The shortest path from $P$ to the circle lies along this normal line. Since the total length of the segment is $CP$, and the circle extends $r$ units from the center toward $P$, we must subtract the radius $r$ to find the distance to the edge of the circle.

Select Answer:

Progressive Hints

Hint 1

Visualize the intersection of the lines

x = 3

and

y = 2

. These lines divide the coordinate plane into four distinct regions. A circle of radius

1

that touches both lines must have its center at a distance of exactly

1

from both lines. Can you visualize the four possible locations for the center of such a circle?

Hint 2

Recall that the perpendicular distance from a point

(h, k)

to a vertical line

x = c

∣ h - c ∣

, and to a horizontal line

y = c

∣ k - c ∣

. Apply this to the lines

x = 3

and

y = 2

with radius

r = 1

. This yields two simple equations:

∣ h - 3∣ = 1

and

∣ k - 2∣ = 1

. Solve these to find the four candidate centers.

Hint 3

Once you identify the four possible centers, calculate the distance of each from the origin

(0, 0)

using the formula

h^{2} + k^{2}

. Select the center that yields the minimum value. Finally, for any point

P (x_{0}, y_{0})

outside a circle with center

(h, k)

and radius

r

, the shortest distance to the circle is given by

(x_{0} - h)^{2} + (y_{0} - k)^{2} - r

Visualized Solution

Visualizing the Point $(3, 2)$

Given point: $(3, 2)$
Lines passing through $(3, 2)$ parallel to axes: $x = 3$ and $y = 2$

Defining the Circle's Center

Let the center of circle $C$ be $(h, k)$ .
Radius $r = 1$ .
Distance from center to $x = 3$ is $1$ .
Distance from center to $y = 2$ is $1$ .

Setting up the Equations

$∣ h - 3∣ = 1$
$∣ k - 2∣ = 1$

Calculating Possible Centers

$h = 3 \pm 1 ⟹ h = 4$ or $h = 2$
$k = 2 \pm 1 ⟹ k = 3$ or $k = 1$
Possible centers: $(4, 3), (2, 3), (4, 1), (2, 1)$

Condition: Closer to the Origin

The circle $C$ is closer to the origin.
We must minimize the distance $h^{2} + k^{2}$ .

Selecting the Correct Center

Distance squared for $(4, 3) = 16 + 9 = 25$
Distance squared for $(2, 3) = 4 + 9 = 13$
Distance squared for $(4, 1) = 16 + 1 = 17$
Distance squared for $(2, 1) = 4 + 1 = 5$
Center of $C$ is $(2, 1)$ .

Locating the Target Point

Target point: $P (5, 5)$
Goal: Find the shortest distance from $P (5, 5)$ to circle $C$ .

Shortest Distance Concept

The shortest distance from a point to a circle lies along the normal.
Normal passes through the center.
Shortest distance $d = C P - r$

Calculating Distance to Center

Center $C (2, 1)$ and Point $P (5, 5)$
$C P = (5 - 2)^{2} + (5 - 1)^{2}$

Evaluating the Distance

$C P = 3^{2} + 4^{2}$
$C P = 9 + 16 = 25 = 5$

Final Shortest Distance

Shortest distance $= C P - r$
Shortest distance $= 5 - 1 = 4$
Final Answer: 4

00:00 / 00:00

The Sigma Insight: Equation of Tangent and Normal

Solution Diagram

Analyzing the Setup

We are given two intersecting lines,

x = 3

and

y = 2

, which act as boundaries on the coordinate plane. We need to place a circle of radius

r = 1

such that it is tangent to both lines.

For a circle with center

(h, k)

to be tangent to the line

x = 3

, the perpendicular distance must satisfy:

∣ h - 3∣ = 1 ⟹ h = 3 \pm 1

Similarly, for the line

y = 2

, the condition is:

∣ k - 2∣ = 1 ⟹ k = 2 \pm 1

This yields four possible centers for the circle:

(4, 3)

(2, 3)

(4, 1)

, and

(2, 1)

Identifying the Optimal Circle

The problem requires the circle to be closest to the origin

(0, 0)

. We calculate the squared distance

d^{2} = h^{2} + k^{2}

for each candidate center:

For $(4, 3)$ : $d^{2} = 4^{2} + 3^{2} = 16 + 9 = 25$ For

(2, 3)

d^{2} = 2^{2} + 3^{2} = 4 + 9 = 13

For $(4, 1)$ : $d^{2} = 4^{2} + 1^{2} = 16 + 1 = 17$ For

(2, 1)

d^{2} = 2^{2} + 1^{2} = 4 + 1 = 5

Comparing these values, the center

(2, 1)

results in the minimum distance to the origin. Thus, our circle

C

has center

(2, 1)

and radius

r = 1

Calculating the Shortest Distance

We must find the shortest distance from the point

P (5, 5)

to the circle

C

. The shortest distance from an external point to a circle is found along the line segment connecting the point to the center of the circle.

First, we calculate the distance

C P

between

P (5, 5)

and the center

C (2, 1)

C P = (5 - 2)^{2} + (5 - 1)^{2}

C P = 3^{2} + 4^{2} = 9 + 16 = 25 = 5

The shortest distance to the boundary of the circle is obtained by subtracting the radius from the total distance

C P

Shortest Distance = C P - r = 5 - 1 = 4

The final shortest distance from the point

P (5, 5)

to the circle

C

is 4.

Common Traps & FAQs

Why are there four possible centers? I thought a circle touching two lines was unique.

That is a fantastic observation! Think of the lines

x = 3

and

y = 2

as a cross. A circle of radius

1

can be placed in any of the four corners created by this cross while still touching both lines. Because the absolute value equations

∣ h - 3∣ = 1

and

∣ k - 2∣ = 1

allow for both positive and negative offsets, we mathematically generate four distinct candidates:

(4, 3), (2, 3), (4, 1),

and

(2, 1)

How do I know which circle is 'closer to the origin'?

The problem explicitly asks for the circle closer to the origin

(0, 0)

. Mathematically, this is a minimization problem. We need to minimize the distance from the origin to the center of the circle, which is defined by the expression

h^{2} + k^{2}

. By calculating this distance for all four candidates, we find that the center

(2, 1)

results in the smallest value,

5

, making it the correct choice.

Why do we subtract the radius at the end?

Imagine drawing a straight line from the external point

P

to the center of the circle

C

. This line segment represents the distance between the point and the center. However, the circle is a boundary, not a solid point. The shortest path from

P

to the circle lies along this normal line. Since the total length of the segment is

C P

, and the circle extends

r

units from the center toward

P

, we must subtract the radius

r

to find the distance to the edge of the circle.

Animated Solution for Mathematics - Circles: Let a circle $C$ of radius 1 and closer to the origin be such that the lines passing through the point $(3, 2)$ and parallel to the coordinate axes touch it. Then the shortest distance of the circle $C$ from the point $(5, 5)$ is :

Select Answer:

Visualized Solution

Visualizing the Point $(3, 2)$

Defining the Circle's Center

Setting up the Equations

Calculating Possible Centers

Condition: Closer to the Origin

Selecting the Correct Center

Locating the Target Point

Shortest Distance Concept

Calculating Distance to Center

Evaluating the Distance

Final Shortest Distance

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Identifying the Optimal Circle

Calculating the Shortest Distance

Similar Questions

Let the lengths of intercepts on $x$ -axis and $y$ -axis made by the circle $x^{2} + y^{2} + a x + 2 a y + c = 0$ ( $a < 0$ ) be $22$ and $25$ , respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line $x + 2 y = 0$ , is equal to :

The straight line $x + 2 y = 1$ meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is :

The line $2 x - y + 1 = 0$ is a tangent to the circle at the point $(2, 5)$ and the centre of the circle lies on $x - 2 y = 4$ . Then, the radius of the circle 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 ________

Two circles with equal radii are intersecting at the points $(0, 1)$ and $(0, - 1)$ . The tangent at the point $(0, 1)$ to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is :

Let the tangent to the circle $x^{2} + y^{2} = 25$ at the point R(3, 4) meet x -axis and y -axis at point P and Q, respectively. If r is the radius of the circle passing through the origin O and having centre at the incentre of the triangle OPQ, then $r^{2}$ is equal to :

Circles with radii 3, 4 and 5 touch each other externally. If $P$ is the point of intersection of tangents to these circles at their points of contact, find the distance of $P$ from the points of contact.

The shortest distance between line $y - x = 1$ and curve $x = y^{2}$ is

Select Answer:

Visualized Solution

Visualizing the Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (3,2)

Defining the Circle's Center

Setting up the Equations

Calculating Possible Centers

Condition: Closer to the Origin

Selecting the Correct Center

Locating the Target Point

Shortest Distance Concept

Calculating Distance to Center

Evaluating the Distance

Final Shortest Distance

The Sigma Insight: Equation of Tangent and Normal

Analyzing the Setup

Identifying the Optimal Circle

Calculating the Shortest Distance

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance between line y−x=1 and curve x=y2 is

Visualizing the Point $(3, 2)$

The line $2 x - y + 1 = 0$ is a tangent to the circle at the point $(2, 5)$ and the centre of the circle lies on $x - 2 y = 4$ . Then, the radius of the circle is:

The shortest distance between line $y - x = 1$ and curve $x = y^{2}$ is