JEE Main 2024 (09 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Circles: Let the centre of a circle, passing through the points $(0, 0), (1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ , be $(h, k)$ . Then for all possible values of the coordinates of the centre $(h, k)$ , $4 (h^{2} + k^{2})$ is equal to

Q: Why do we reject the case $r = 3 + r$?

If we set $r = 3 + r$, we get $0 = 3$, which is a mathematical contradiction. Physically, this means the circle cannot touch the larger circle externally while passing through the origin, because the radius would have to be infinite to reach that far. It is a beautiful example of how algebra acts as a guardrail for our physical intuition.

Q: Can I solve this using the general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$?

Absolutely! You can substitute the points $(0,0)$ and $(1,0)$ to find $c$ and $g$, then use the tangency condition. However, that path is algebraically heavy and prone to calculation errors. The geometric approach we used is the 'JEE way'—it leverages symmetry and properties to bypass tedious algebra, saving you precious time in the exam hall.

Enter Numerical Value:

Visualized Solution

Visualizing the Given Elements

Given circle: $x^{2} + y^{2} = 9$
Center $C_{1} (0, 0)$ , Radius $R = 3$
Required circle passes through $O (0, 0)$ and $A (1, 0)$

The Perpendicular Bisector Theorem

The center $(h, k)$ of any circle passing through two points lies on their perpendicular bisector.
Chord endpoints: $(0, 0)$ and $(1, 0)$

Locating the Center's x-coordinate

Midpoint of $(0, 0)$ and $(1, 0)$ is $(\frac{1}{2}, 0)$
Perpendicular bisector equation: $x = \frac{1}{2}$
Therefore, the x-coordinate of the center is $h = \frac{1}{2}$

Defining the Radius $r$

Let the radius of the required circle be $r$ .
Distance from center $(h, k)$ to a point on the circle $(0, 0)$ is $r$ .
$r = (h - 0)^{2} + (k - 0)^{2} = h^{2} + k^{2}$

Distance Between Centers

Center of given circle: $C_{1} (0, 0)$
Center of required circle: $C_{2} (h, k)$
Distance between centers $d = h^{2} + k^{2}$
Notice that $d$ is exactly equal to $r$ !

The Condition for Touching Circles

For two circles to touch each other, the distance between their centers $d$ must relate to their radii.
General condition: $d = ∣ R \pm r ∣$
Where $R$ is the radius of the big circle, and $r$ is the radius of the small circle.

Applying the Touching Condition

Substitute $d = r$
Substitute $R = 3$
The condition becomes: $r = ∣3 \pm r ∣$

Solving for $r$

Case 1 (External touch): $r = 3 + r ⟹ 0 = 3$ (Impossible)
Case 2 (Internal touch): $r = ∣3 - r ∣$
Since $r > 0$ , we take $r = 3 - r$
$2 r = 3 ⟹ r = \frac{3}{2}$

Calculating $h^{2} + k^{2}$

We know $r = h^{2} + k^{2}$
Squaring both sides: $h^{2} + k^{2} = r^{2}$
Substitute $r = \frac{3}{2}$ :
$h^{2} + k^{2} = (\frac{3}{2})^{2} = \frac{9}{4}$

The Final Answer

The question asks for the value of $4 (h^{2} + k^{2})$
$4 (h^{2} + k^{2}) = 4 \times \frac{9}{4}$
$4 (h^{2} + k^{2}) = 9$

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The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

Imagine a fixed circle centered at the origin

(0, 0)

with a radius of

R = 3

. We are tasked with finding the center

(h, k)

of a new circle that passes through the points

(0, 0)

and

(1, 0)

while touching the fixed circle.

The Perpendicular Bisector

The chord connecting

(0, 0)

and

(1, 0)

must have its perpendicular bisector passing through the center

(h, k)

of the new circle. The midpoint of this chord is

(\frac{1}{2}, 0)

Since the chord lies on the x-axis, the perpendicular bisector is the vertical line:

h = \frac{1}{2}

The Radius-Distance Duality

Let

r

be the radius of the new circle. Since the circle passes through the origin

(0, 0)

, the distance from the center

(h, k)

to the origin is equal to

r

Using the distance formula:

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

Note that the distance

d

between the center of the fixed circle

(0, 0)

and the center of the new circle

(h, k)

is also

h^{2} + k^{2}

. Therefore, we have the identity:

d = r

The Touching Condition

For two circles with radii

R

and

r

and distance

d

between their centers to touch, the condition is

d = ∣ R \pm r ∣

. Given

R = 3

and

d = r

, we substitute these values:

r = ∣3 \pm r ∣

If we assume external touching,

r = 3 + r

, which leads to

0 = 3

, an impossibility. Thus, the circles must touch internally.

We solve for

r

using the internal touching condition:

r = ∣3 - r ∣

r = 3 - r

2 r = 3 \Rightarrow r = \frac{3}{2}

Final Calculation

We have established that

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

. Substituting the value of

r

h^{2} + k^{2} = (\frac{3}{2})^{2} = \frac{9}{4}

The problem asks for the value of

4 (h^{2} + k^{2})

. Performing the final arithmetic:

4 \times (\frac{9}{4}) = 9

The final result is 9.

Common Traps & FAQs

Why do we reject the case

r = 3 + r

If we set

r = 3 + r

, we get

0 = 3

, which is a mathematical contradiction. Physically, this means the circle cannot touch the larger circle externally while passing through the origin, because the radius would have to be infinite to reach that far. It is a beautiful example of how algebra acts as a guardrail for our physical intuition.

Can I solve this using the general equation of a circle

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

Absolutely! You can substitute the points

(0, 0)

and

(1, 0)

to find

c

and

g

, then use the tangency condition. However, that path is algebraically heavy and prone to calculation errors. The geometric approach we used is the 'JEE way'—it leverages symmetry and properties to bypass tedious algebra, saving you precious time in the exam hall.

Is it always true that

d = r

Only in this specific problem! Because the new circle passes through the origin

(0, 0)

, and the fixed circle is centered at the origin

(0, 0)

, the distance between their centers is exactly the radius of the new circle. Always look for these 'coincidences'—they are usually the key to the problem.

Animated Solution for Mathematics - Circles: Let the centre of a circle, passing through the points $(0, 0), (1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ , be $(h, k)$ . Then for all possible values of the coordinates of the centre $(h, k)$ , $4 (h^{2} + k^{2})$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Given Elements

The Perpendicular Bisector Theorem

Locating the Center's x-coordinate

Defining the Radius $r$

Distance Between Centers

The Condition for Touching Circles

Applying the Touching Condition

Solving for $r$

Calculating $h^{2} + k^{2}$

The Final Answer

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Perpendicular Bisector

The Radius-Distance Duality

The Touching Condition

Final Calculation

Similar Questions

The centre of the circle passing through $(0, 0)$ and $(1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ is

The centre of a circle passing through the points $(0, 0), (1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ is

A circle is given by $x^{2} + (y - 1)^{2} = 1$ , another circle $C$ touches it externally and also the x-axis, then the locus of its centre is

Let a circle $C$ touch the lines $L_{1} : 4 x - 3 y + K_{1} = 0$ and $L_{2} : 4 x - 3 y + K_{2} = 0, K_{1}, K_{2} \in R$ . If a line passing through the centre of the circle $C$ intersects $L_{1}$ at $(- 1, 2)$ and $L_{2}$ at $(3, - 6)$ , then the equation of the circle $C$ is

The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length $3 a$ is

The locus of the centres of the circles, which touch the circle, $x^{2} + y^{2} = 1$ externally, also touch the y-axis and lie in the first quadrant, is :

If the four distinct points $(4, 6), (- 1, 5), (0, 0)$ and $(k, 3 k)$ lie on a circle of radius $r$ , then $10 k + r^{2}$ is equal to

Let $C : x^{2} + y^{2} = 4$ and $C^{'} : x^{2} + y^{2} - 4 λ x + 9 = 0$ be two circles. If the set of all values of $λ$ so that the circles $C$ and $C^{'}$ intersect at two distinct points, is $R - [a, b]$ , then the point $(8 a + 12, 16 b - 20)$ lies on the curve :

The two circles $x^{2} + y^{2} = a x$ and $x^{2} + y^{2} = c^{2}$ ( $c > 0$ ) touch each other if

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Enter Numerical Value:

Visualized Solution

Visualizing the Given Elements

The Perpendicular Bisector Theorem

Locating the Center's x-coordinate

Defining the Radius .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Distance Between Centers

The Condition for Touching Circles

Applying the Touching Condition

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

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h2+k2

The Final Answer

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

The Perpendicular Bisector

The Radius-Distance Duality

The Touching Condition

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The centre of the circle passing through (0,0) and (1,0) and touching the circle x2+y2=9 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The centre of a circle passing through the points (0,0),(1,0) and touching the circle x2+y2=9 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A circle is given by x2+(y−1)2=1, another circle C touches it externally and also the x-axis, then the locus of its centre is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the centres of the circles, which touch the circle, x2+y2=1 externally, also touch the y-axis and lie in the first quadrant, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the four distinct points (4,6),(−1,5),(0,0) and (k,3k) lie on a circle of radius r, then 10k+r2 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The two circles x2+y2=ax and x2+y2=c2 (c>0) touch each other if

Defining the Radius $r$

Solving for $r$

Calculating $h^{2} + k^{2}$

The centre of the circle passing through $(0, 0)$ and $(1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ is

The centre of a circle passing through the points $(0, 0), (1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ is

A circle is given by $x^{2} + (y - 1)^{2} = 1$ , another circle $C$ touches it externally and also the x-axis, then the locus of its centre is

The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length $3 a$ is

The locus of the centres of the circles, which touch the circle, $x^{2} + y^{2} = 1$ externally, also touch the y-axis and lie in the first quadrant, is :

If the four distinct points $(4, 6), (- 1, 5), (0, 0)$ and $(k, 3 k)$ lie on a circle of radius $r$ , then $10 k + r^{2}$ is equal to

The two circles $x^{2} + y^{2} = a x$ and $x^{2} + y^{2} = c^{2}$ ( $c > 0$ ) touch each other if