JEE Main 2022 (27 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Circles: The set of values of $k$ for which the circle $C : 4 x^{2} + 4 y^{2} - 12 x + 8 y + k = 0$ lies inside the fourth quadrant and the point $(1, - \frac{1}{3})$ lies on or inside the circle $C$ is :

Q: Why do we use $r < |x_c|$ and $r < |y_c|$? Shouldn't it be $\le$?

Great observation! If $r = |x_c|$, the circle would be tangent to the $y$-axis, meaning it would touch the boundary of the quadrant. The problem states it lies *inside* the fourth quadrant, implying it shouldn't cross or touch the axes. Thus, strict inequality is the safer, more precise choice.

Q: Why do we check $S(1, -\frac{1}{3}) \le 0$?

The expression $S(x, y) = x^2 + y^2 + 2gx + 2fy + c$ represents the power of a point. If $S < 0$, the point is inside. If $S = 0$, it's on the boundary. Since the problem says 'on or inside', we use $\le 0$.

Q: Why did we divide by 4 at the start?

Standard form is the key to unlocking the circle's secrets. By dividing by 4, we normalize the coefficients of $x^2$ and $y^2$ to 1, which allows us to instantly identify the center and radius using the standard form $x^2 + y^2 + 2gx + 2fy + c = 0$.

Select Answer:

Progressive Hints

Hint 1

Visualize the circle trapped in the fourth quadrant. If the circle is entirely in the fourth quadrant, it cannot cross the

y

-axis (where

x = 0

) or the

x

-axis (where

y = 0

). This implies that the distance from the center to each axis must be strictly greater than the radius. Can you write down the center

(h, k)

and radius

r

in terms of

k

Hint 2

To ensure the circle stays in the fourth quadrant, you need two inequalities: the distance from the center to the

y

-axis must be greater than the radius, and the distance from the center to the

x

-axis must be greater than the radius. Also, remember the condition for a point

(x_{1}, y_{1})

to lie inside or on a circle

S (x, y) = 0

S (x_{1}, y_{1}) \leq 0

Hint 3

Start by normalizing the equation:

x^{2} + y^{2} - 3 x + 2 y + \frac{k}{4} = 0

. The center is

(\frac{3}{2}, - 1)

. The radius is

r = (\frac{3}{2})^{2} + (- 1)^{2} - \frac{k}{4}

. Now, apply the constraints:

r < \frac{3}{2}

(for the

y

-axis) and

r < ∣ - 1∣

(for the

x

-axis), and finally, substitute the point

(1, - \frac{1}{3})

into the circle equation.

Visualized Solution

Visualizing the Problem

Given equation: $4 x^{2} + 4 y^{2} - 12 x + 8 y + k = 0$
Must lie entirely in the $4^{th}$ quadrant.
Point $(1, - \frac{1}{3})$ lies on or inside $C$ .

Standardizing the Circle Equation

Given: $4 x^{2} + 4 y^{2} - 12 x + 8 y + k = 0$
Divide by $4$ to get standard form $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .
Standard form: $x^{2} + y^{2} - 3 x + 2 y + \frac{k}{4} = 0$

Finding Center and Radius

Compare with $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
Center $C (- g, - f) = (\frac{3}{2}, - 1)$
Radius $r = g^{2} + f^{2} - c = (- \frac{3}{2})^{2} + (1)^{2} - \frac{k}{4}$
Simplified radius: $r = \frac{13 - k}{2}$

Condition for a Real Circle

For a circle to physically exist, its radius must be real and positive.
$13 - k > 0$
Result: $k < 13$

Constraint: Staying Right of the $y$ -axis

For the circle to be entirely in the $4^{th}$ quadrant, it must not cross the $y$ -axis.
Distance from center to $y$ -axis must be greater than radius.
$x_{c} > r ⟹ \frac{3}{2} > \frac{13 - k}{2}$

Solving the $y$ -axis Constraint

$\frac{3}{2} > \frac{13 - k}{2}$
$3 > 13 - k$
Squaring both sides: $9 > 13 - k$
Result: $k > 4$

Constraint: Staying Below the $x$ -axis

The circle must also not cross the $x$ -axis.
Distance from center to $x$ -axis must be greater than radius.
$∣ y_{c} ∣ > r ⟹ ∣ - 1∣ > \frac{13 - k}{2}$

Solving the $x$ -axis Constraint

$1 > \frac{13 - k}{2}$
$2 > 13 - k$
Squaring both sides: $4 > 13 - k$
Result: $k > 9$

Combining Quadrant Constraints

From $y$ -axis constraint: $k > 4$
From $x$ -axis constraint: $k > 9$
From existence constraint: $k < 13$
Combined range so far: $k \in (9, 13)$

Point Position Constraint

Point $P (1, - \frac{1}{3})$ lies on or inside the circle.
Geometric condition: $S_{1} \leq 0$
Substitute $P$ into the original circle equation $S$ .

Substituting the Point

Original equation: $S = 4 x^{2} + 4 y^{2} - 12 x + 8 y + k$
Substitute $x = 1, y = - \frac{1}{3}$ :
$4 (1)^{2} + 4 (- \frac{1}{3})^{2} - 12 (1) + 8 (- \frac{1}{3}) + k \leq 0$

Evaluating the Inequality

$4 (1) + 4 (\frac{1}{9}) - 12 - \frac{8}{3} + k \leq 0$
$4 + \frac{4}{9} - 12 - \frac{24}{9} + k \leq 0$
$- 8 - \frac{20}{9} + k \leq 0$

Solving for $k$

$- 8 - \frac{20}{9} + k \leq 0$
$\frac{- 72 - 20}{9} + k \leq 0$
$\frac{- 92}{9} + k \leq 0$
Result: $k \leq \frac{92}{9}$

Final Intersection

From quadrant constraints: $k \in (9, 13)$
From point constraint: $k \leq \frac{92}{9}$
Note: $\frac{92}{9} \approx 10.22$ , which is less than $13$ .
Final Intersection: $k \in (9, \frac{92}{9}]$

00:00 / 00:00

The Sigma Insight: Position of a Point with Respect to a Circle

Solution Diagram

Analyzing the Setup

We begin with the given equation of the circle:

4 x^{2} + 4 y^{2} - 12 x + 8 y + k = 0

To bring order to this expression, we divide the entire equation by

4

to obtain the standard form:

x^{2} + y^{2} - 3 x + 2 y + \frac{k}{4} = 0

By comparing this to the general form

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

, we identify the center of the circle at

(\frac{3}{2}, - 1)

. The radius

r

is calculated as follows:

r = (\frac{3}{2})^{2} + (- 1)^{2} - \frac{k}{4} = \frac{13 - k}{2}

For the circle to exist, the radius must be a real, positive value. This imposes our first constraint:

13 - k > 0 \Rightarrow k < 13

Geometric Confinement

For the circle to remain strictly within the fourth quadrant, it must not intersect the

x

-axis (

y = 0

) or the

y

-axis (

x = 0

). This requires the distance from the center to each axis to be strictly greater than the radius.

For the

y

-axis, the distance is the absolute value of the

x

-coordinate:

\frac{3}{2} > \frac{13 - k}{2} \Rightarrow 9 > 13 - k \Rightarrow k > 4

For the

x

-axis, the distance is the absolute value of the

y

-coordinate:

1 > \frac{13 - k}{2} \Rightarrow 2 > 13 - k \Rightarrow 4 > 13 - k \Rightarrow k > 9

Combining these constraints, we establish that

k

must lie in the interval

(9, 13)

The Point Constraint

We are given that the point

(1, - \frac{1}{3})

must lie inside or on the boundary of the circle. We substitute this point into the circle equation

S (x, y) = 4 x^{2} + 4 y^{2} - 12 x + 8 y + k \leq 0

4 (1)^{2} + 4 (- \frac{1}{3})^{2} - 12 (1) + 8 (- \frac{1}{3}) + k \leq 0

Simplifying the arithmetic:

4 + \frac{4}{9} - 12 - \frac{8}{3} + k \leq 0

- 8 - \frac{20}{9} + k \leq 0 \Rightarrow k \leq \frac{92}{9}

Final Calculation

By synthesizing the geometric confinement (

k > 9

) and the point inclusion constraint (

k \leq \frac{92}{9}

), we arrive at the final range for

k

The valid interval for

k

is: $9 < k \leq \frac{92}{9}$

Common Traps & FAQs

Why do we use

r < ∣ x_{c} ∣

and

r < ∣ y_{c} ∣

? Shouldn't it be

\leq

Great observation! If

r = ∣ x_{c} ∣

, the circle would be tangent to the

y

-axis, meaning it would touch the boundary of the quadrant. The problem states it lies *inside* the fourth quadrant, implying it shouldn't cross or touch the axes. Thus, strict inequality is the safer, more precise choice.

Why do we check

S (1, - \frac{1}{3}) \leq 0

The expression

S (x, y) = x^{2} + y^{2} + 2 g x + 2 f y + c

represents the power of a point. If

S < 0

, the point is inside. If

S = 0

, it's on the boundary. Since the problem says 'on or inside', we use

\leq 0

Why did we divide by 4 at the start?

Standard form is the key to unlocking the circle's secrets. By dividing by 4, we normalize the coefficients of

x^{2}

and

y^{2}

to 1, which allows us to instantly identify the center and radius using the standard form

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

Animated Solution for Mathematics - Circles: The set of values of $k$ for which the circle $C : 4 x^{2} + 4 y^{2} - 12 x + 8 y + k = 0$ lies inside the fourth quadrant and the point $(1, - \frac{1}{3})$ lies on or inside the circle $C$ is :

Select Answer:

Visualized Solution

Visualizing the Problem

Standardizing the Circle Equation

Finding Center and Radius

Condition for a Real Circle

Constraint: Staying Right of the $y$ -axis

Solving the $y$ -axis Constraint

Constraint: Staying Below the $x$ -axis

Solving the $x$ -axis Constraint

Combining Quadrant Constraints

Point Position Constraint

Substituting the Point

Evaluating the Inequality

Solving for $k$

Final Intersection

The Sigma Insight: Position of a Point with Respect to a Circle

Analyzing the Setup

Geometric Confinement

The Point Constraint

Final Calculation

Similar Questions

If a variable line, $3 x + 4 y - λ = 0$ is such that the two circles $x^{2} + y^{2} - 2 x - 2 y + 1 = 0$ and $x^{2} + y^{2} - 18 x - 2 y + 78 = 0$ are on its opposite sides, then the set of all values of $λ$ is the interval :-

The straight line $2 x - 3 y = 1$ divides the circular region $x^{2} + y^{2} \leq 6$ into two parts. If $S = {(2, 3/4), (5/2, 3/4), (1/4, - 1/4), (1/8, 1/4)}$ then the number of points (s) in $S$ lying inside the smaller part is

Let the circle $S : 36 x^{2} + 36 y^{2} - 108 x + 120 y + C = 0$ be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, $x - 2 y = 4$ and $2 x - y = 5$ lies inside the circle $S$ , then :

Let $r_{1}$ and $r_{2}$ be the radii of the largest and smallest circles, respectively, which pass through the point $(- 4, 1)$ and having their centres on the circumference of the circle $x^{2} + y^{2} + 2 x + 4 y - 4 = 0$ . If $\frac{r _{1}}{r _{2}} = a + b 2$ , then $a + b$ is equal to :

Comprehension Passage

Consider $M$ with $r = \frac{1025}{513}$ . Let $k$ be the number of all those circles $C_{n}$ that are inside $M$ . Let $l$ be the maximum possible number of circles among these $k$ circles such that no two circles intersect. Then

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Let the maximum and minimum values of $(8 x - x^{2} - 12 - 4)^{2} + (x - 7)^{2}, x \in R$ be $M$ and $m$ , respectively. Then $M^{2} - m^{2}$ is equal to _________

Select Answer:

Visualized Solution

Visualizing the Problem

Standardizing the Circle Equation

Finding Center and Radius

Condition for a Real Circle

Constraint: Staying Right of the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-axis

Solving the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-axis Constraint

Constraint: Staying Below the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-axis

Solving the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-axis Constraint

Combining Quadrant Constraints

Point Position Constraint

Substituting the Point

Evaluating the Inequality

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

Final Intersection

The Sigma Insight: Position of a Point with Respect to a Circle

Analyzing the Setup

Geometric Confinement

The Point Constraint

Final Calculation

Similar Questions

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Consider M with r=2198(2199−1)2​​. The number of all those circles Dn​ that are inside M is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Choose the incorrect statement about the two circles whose equations are given below: x2+y2−10x−10y+41=0 and x2+y2−16x−10y+80=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the maximum and minimum values of (8x−x2−12​−4)2+(x−7)2,x∈R be M and m, respectively. Then M2−m2 is equal to _________

Constraint: Staying Right of the $y$ -axis

Solving the $y$ -axis Constraint

Constraint: Staying Below the $x$ -axis

Solving the $x$ -axis Constraint

Solving for $k$

Consider $M$ with $r = \frac{( 2 ^{199} - 1 ) 2}{2 ^{198}}$ . The number of all those circles $D_{n}$ that are inside $M$ is

Choose the incorrect statement about the two circles whose equations are given below: $x^{2} + y^{2} - 10 x - 10 y + 41 = 0$ and $x^{2} + y^{2} - 16 x - 10 y + 80 = 0$

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