Why did we use $\frac{22}{7}$ for $\pi$?

It is a common convention in competitive exams when the area is a multiple of $22$ or $11$. Since $154$ is $22 \times 7$, using $\frac{22}{7}$ makes the calculation $r^2 = 154 \times \frac{7}{22}$ very clean, leading directly to $r^2 = 49$.

JEE Main 2003

LEVELJEE Main

Animated Solution for Mathematics - Circles: The lines $2 x - 3 y = 5$ and $3 x - 4 y = 7$ are diameters of a circle having area as 154 sq.units. Then the equation of the circle is

Q: Can I use substitution instead of elimination to find the center?

Absolutely! You can solve $2x - 3y = 5$ for $x$ as $x = \frac{5 + 3y}{2}$ and substitute this into the second equation. Both methods are perfectly valid; choose the one that feels more comfortable for your algebraic workflow.

Select Answer:

Visualized Solution

Visualizing the Circle

Given diameters: $2 x - 3 y = 5$ and $3 x - 4 y = 7$
Area of the circle = $154$ sq. units

The Intersection Property

The intersection of any two diameters is the center of the circle.
Let the center be $(h, k)$ .

System of Equations

$2 x - 3 y = 5$ ... (1)
$3 x - 4 y = 7$ ... (2)

Solving for $x$

Multiply (1) by $4$ : $8 x - 12 y = 20$
Multiply (2) by $3$ : $9 x - 12 y = 21$
Subtracting: $(9 x - 8 x) = 21 - 20 \Rightarrow x = 1$

Solving for $y$

Substitute $x = 1$ in (1):
$2 (1) - 3 y = 5$
$- 3 y = 3 \Rightarrow y = - 1$
Center $(h, k) = (1, - 1)$

Finding the Radius

Area of circle = $π r^{2} = 154$
Using $π = \frac{22}{7}$

Calculating $r^{2}$

$\frac{22}{7} r^{2} = 154$
$r^{2} = 154 \times \frac{7}{22}$
$r^{2} = 7 \times 7 = 49$

Standard Equation Form

Equation: $(x - h)^{2} + (y - k)^{2} = r^{2}$
Substitute $h = 1, k = - 1, r^{2} = 49$ :
$(x - 1)^{2} + (y - (- 1))^{2} = 49$

Expanding the Equation

$(x - 1)^{2} + (y + 1)^{2} = 49$
Expanding squares:
$(x^{2} - 2 x + 1) + (y^{2} + 2 y + 1) = 49$

Final General Equation

Combine constant terms: $1 + 1 = 2$
$x^{2} + y^{2} - 2 x + 2 y + 2 = 49$
Final form: $x^{2} + y^{2} - 2 x + 2 y = 47$

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The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Analyzing the Setup

In the coordinate plane, a circle is defined by its center

(h, k)

and its radius

r

. We are given two diameters,

2 x - 3 y = 5

and

3 x - 4 y = 7

Since every diameter of a circle must pass through its center, the intersection point of these two lines is the center of the circle. We solve the system of equations:

2 x - 3 y = 5

3 x - 4 y = 7

Finding the Center

To solve this system, we multiply the first equation by

4

and the second by

3

to align the coefficients of

y

8 x - 12 y = 20

9 x - 12 y = 21

Subtracting the first equation from the second yields

x = 1

. Substituting

x = 1

into the first equation:

2 (1) - 3 y = 5

- 3 y = 3

y = - 1

Thus, the center of the circle is

(h, k) = (1, - 1)

Unlocking the Radius

The area of the circle is given as

154

square units. Using the formula

A = π r^{2}

and the approximation

π = \frac{22}{7}

, we set up the following:

\frac{22}{7} r^{2} = 154

Solving for

r^{2}

r^{2} = 154 \times \frac{7}{22}

r^{2} = 7 \times 7 = 49

The Final Synthesis

The standard form of a circle's equation is

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

. Substituting our center

(1, - 1)

and

r^{2} = 49

(x - 1)^{2} + (y + 1)^{2} = 49

Expanding the binomials:

(x^{2} - 2 x + 1) + (y^{2} + 2 y + 1) = 49

x^{2} + y^{2} - 2 x + 2 y + 2 = 49

Subtracting

2

from both sides, we obtain the final equation of the circle:

$x^{2} + y^{2} - 2 x + 2 y = 47$

Common Traps & FAQs

Wait, why do we assume the intersection of these two lines is the center?

That is a fundamental property of a circle! By definition, a diameter is a chord that passes through the center. Since every diameter must pass through the center, any two distinct diameters must intersect at that exact point. It is the only point they share.

Can I use substitution instead of elimination to find the center?

Absolutely! You can solve

2 x - 3 y = 5

for

x

x = \frac{5 + 3 y}{2}

and substitute this into the second equation. Both methods are perfectly valid; choose the one that feels more comfortable for your algebraic workflow.

Why did we use

\frac{22}{7}

for

π

It is a common convention in competitive exams when the area is a multiple of

22

11

. Since

154

22 \times 7

, using

\frac{22}{7}

makes the calculation

r^{2} = 154 \times \frac{7}{22}

very clean, leading directly to

r^{2} = 49

Animated Solution for Mathematics - Circles: The lines $2 x - 3 y = 5$ and $3 x - 4 y = 7$ are diameters of a circle having area as 154 sq.units. Then the equation of the circle is

Select Answer:

Visualized Solution

Visualizing the Circle

The Intersection Property

System of Equations

Solving for $x$

Solving for $y$

Finding the Radius

Calculating $r^{2}$

Standard Equation Form

Expanding the Equation

Final General Equation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

Finding the Center

Unlocking the Radius

The Final Synthesis

Similar Questions

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Animated Solution for Mathematics - Circles: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The lines 2x−3y=5 and 3x−4y=7 are diameters of a circle having area as 154 sq.units. Then the equation of the circle is

Select Answer:

Visualized Solution

Visualizing the Circle

The Intersection Property

System of Equations

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

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

Finding the Radius

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r2

Standard Equation Form

Expanding the Equation

Final General Equation

The Sigma Insight: Standard and General Equation of a Circle

Analyzing the Setup

Finding the Center

Unlocking the Radius

The Final Synthesis

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Solving for $x$

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Calculating $r^{2}$

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