JEE Main 2021 (18 March Shift 1)

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Animated Solution for Mathematics - Circles: For the four circles $M, N, O$ and $P$ , following four equations are given : Circle $M : x^{2} + y^{2} = 1$ Circle $N : x^{2} + y^{2} - 2 x = 0$ Circle $O : x^{2} + y^{2} - 2 x - 2 y + 1 = 0$ Circle $P : x^{2} + y^{2} - 2 y = 0$ If the centre of circle $M$ is joined with centre of the circle $N$ , further centre of circle $N$ is joined with centre of the circle $O$ , centre of circle $O$ is joined with the centre of circle $P$ and lastly, centre of circle $P$ is joined with centre of circle $M$ , then these lines form the sides of a :

Q: Is there a faster way to find the center of a circle without writing out the full general equation every time?

Absolutely! For an equation like $x^2 + y^2 + 2gx + 2fy + c = 0$, the center is simply $(-g, -f)$. Just take the coefficient of $x$, divide by $-2$, and take the coefficient of $y$, divide by $-2$. For example, in $x^2 + y^2 - 2x = 0$, the $x$-coefficient is $-2$. Divide by $-2$ to get $1$. The $y$-coefficient is $0$. Divide by $-2$ to get $0$. Center is $(1,0)$. Fast, right?

Select Answer:

Progressive Hints

Hint 1

The Spark: Do not let the algebraic equations intimidate you. In coordinate geometry, every circle equation is merely a mask for a simple point in the Cartesian plane. Your first mission is to strip away that mask. Use the general form

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

to extract the center

(h, k)

for each circle. Once you have these four coordinates, plot them mentally or on a scratchpad. You will immediately see the skeleton of the shape.

Hint 2

The Arsenal: Once you have the coordinates, you are playing the game of distance. Recall the distance formula

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

. To distinguish between a rhombus and a square, you need more than just side lengths. A rhombus has equal sides, but a square demands equal diagonals too. You must arm yourself with both calculations to be certain.

Hint 3

The Execution: Start by calculating the distance between the center of

M

(0, 0)

and

N

(1, 0)

. Then, calculate the distance between

N

(1, 0)

and

O

(1, 1)

. If you find the side lengths are all equal to

1

, do not stop! Calculate the length of the diagonal connecting

M

(0, 0)

and

O

(1, 1)

using

(1 - 0)^{2} + (1 - 0)^{2}

. If the diagonal is

2

, you have your answer.

Visualized Solution

Analyze Circle $M$

Circle $M : x^{2} + y^{2} = 1$
Standard form: $(x - h)^{2} + (y - k)^{2} = r^{2}$
Center $C_{M} = (0, 0)$
Radius $r = 1$

Analyze Circle $N$

Circle $N : x^{2} + y^{2} - 2 x = 0$
General form: $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
Comparing coefficients: $2 g = - 2 ⟹ g = - 1$
Center $C_{N} = (- g, - f) = (1, 0)$

Analyze Circle $O$

Circle $O : x^{2} + y^{2} - 2 x - 2 y + 1 = 0$
Comparing coefficients: $2 g = - 2$ and $2 f = - 2$
This gives $g = - 1$ and $f = - 1$
Center $C_{O} = (1, 1)$

Analyze Circle $P$

Circle $P : x^{2} + y^{2} - 2 y = 0$
Comparing coefficients: $2 g = 0$ and $2 f = - 2$
This gives $g = 0$ and $f = - 1$
Center $C_{P} = (0, 1)$

Join the Centers

Join the centers in order: $M \to N \to O \to P \to M$
Vertices of the quadrilateral are $(0, 0), (1, 0), (1, 1),$ and $(0, 1)$

Calculate Side Lengths

Calculate side lengths using distance formula:
$M N = (1 - 0)^{2} + (0 - 0)^{2} = 1$
$N O = (1 - 1)^{2} + (1 - 0)^{2} = 1$
$O P = (0 - 1)^{2} + (1 - 1)^{2} = 1$
$P M = (0 - 0)^{2} + (0 - 1)^{2} = 1$
All four sides are equal to $1$ unit.

Check Diagonals

Check the lengths of the diagonals:
$M O = (1 - 0)^{2} + (1 - 0)^{2} = 2$
$N P = (0 - 1)^{2} + (1 - 0)^{2} = 2$
Both diagonals are equal to $2$ .

Final Conclusion

Since all sides are equal ( $1$ unit) and diagonals are equal ( $2$ units):
The quadrilateral is a Square.
Final Answer: Square

00:00 / 00:00

The Sigma Insight: Standard and General Equation of a Circle

Solution Diagram

Decoding the Circle Centers

We begin by identifying the centers of the four given circles. The general equation of a circle is

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

, where the center is

(- g, - f)

For Circle

M

(

x^{2} + y^{2} = 1

), the center

C_{M}

(0, 0)

For Circle

N

(

x^{2} + y^{2} - 2 x = 0

), we identify

2 g = - 2

, so

g = - 1

and

f = 0

. The center

C_{N}

(1, 0)

For Circle

O

(

x^{2} + y^{2} - 2 x - 2 y + 1 = 0

), we identify

2 g = - 2

and

2 f = - 2

, yielding

g = - 1

and

f = - 1

. The center

C_{O}

(1, 1)

For Circle

P

(

x^{2} + y^{2} - 2 y = 0

), we identify

g = 0

and

2 f = - 2

, so

f = - 1

. The center

C_{P}

(0, 1)

Analyzing the Quadrilateral Sides

We now examine the quadrilateral formed by the vertices

C_{M} (0, 0)

C_{N} (1, 0)

C_{O} (1, 1)

, and

C_{P} (0, 1)

. We use the distance formula

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

to find the side lengths:

M N = (1 - 0)^{2} + (0 - 0)^{2} = 1

N O = (1 - 1)^{2} + (1 - 0)^{2} = 1

O P = (0 - 1)^{2} + (1 - 1)^{2} = 1

P M = (0 - 0)^{2} + (0 - 1)^{2} = 1

Since all sides are equal to

1

, the quadrilateral is a rhombus.

Verifying the Geometry

To determine if the shape is a square, we must compare the lengths of the diagonals

M O

and

N P

For diagonal

M O

connecting

(0, 0)

and

(1, 1)

M O = (1 - 0)^{2} + (1 - 0)^{2} = 1 + 1 = 2

For diagonal

N P

connecting

(1, 0)

and

(0, 1)

N P = (0 - 1)^{2} + (1 - 0)^{2} = 1 + 1 = 2

Because all sides are equal and both diagonals are equal, the quadrilateral

M N O P

is a square.

Common Traps & FAQs

Wait, if I calculate all four sides and they are equal, why isn't it automatically a square? Why do I have to check the diagonals?

That is a brilliant question! In geometry, a rhombus is a quadrilateral with all four sides equal. A square is a special type of rhombus where the internal angles are

9 0^{\circ}

. If you only check the sides, you have only proven it is a rhombus. To confirm it is a square, you must either check that the diagonals are equal in length or that the adjacent sides are perpendicular (i.e., their slopes multiply to

- 1

Is there a faster way to find the center of a circle without writing out the full general equation every time?

Absolutely! For an equation like

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

, the center is simply

(- g, - f)

. Just take the coefficient of

x

, divide by

- 2

, and take the coefficient of

y

, divide by

- 2

. For example, in

x^{2} + y^{2} - 2 x = 0

, the

x

-coefficient is

- 2

. Divide by

- 2

to get

1

. The

y

-coefficient is

0

. Divide by

- 2

to get

0

. Center is

(1, 0)

. Fast, right?

I keep getting confused with the signs when finding the center. How do I avoid silly mistakes?

The sign trap is the most common pitfall in JEE! Always rewrite the equation in the form

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

if you are unsure. If you use the

(- g, - f)

method, write down the coefficients clearly first. For

x^{2} + y^{2} - 2 x - 2 y + 1 = 0

, write

2 g = - 2

and

2 f = - 2

. Then solve for

g

and

f

explicitly before negating them. Precision is your best friend.

Select Answer:

Visualized Solution

Analyze Circle $M$

Analyze Circle $N$

Analyze Circle $O$

Analyze Circle $P$

Join the Centers

Calculate Side Lengths

Check Diagonals

Final Conclusion

The Sigma Insight: Standard and General Equation of a Circle

Decoding the Circle Centers

Analyzing the Quadrilateral Sides

Verifying the Geometry

Similar Questions

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

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 :

If $m_{i}, \frac{1}{m _{i}}, m_{i} > 0, i = 1, 2, 3, 4$ are four distinct points on a circle, then show that $m_{1} m_{2} m_{3} m_{4} = 1$ .

Let $A B C D$ and $A E F G$ be squares of side 4 and 2 units, respectively. The point $E$ is on the line segment $A B$ and the point $F$ is on the diagonal $A C$ . Then the radius $r$ of the circle passing through the point $F$ and touching the line segments $B C$ and $C D$ satisfies:

If the lines $2 x + 3 y + 1 = 0$ and $3 x - y - 4 = 0$ lie along diameter of a circle of circumference $10 π$ , then the equation of the circle 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 :

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 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 :

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 $P$ and $Q$ be two distinct points on a circle which has center at $C (2, 3)$ and which passes through origin $O$ . If $O C$ is perpendicular to both the line segments $C P$ and $C Q$ , then the set ${P, Q}$ is equal to

Select Answer:

Visualized Solution

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

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

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

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

Join the Centers

Calculate Side Lengths

Check Diagonals

Final Conclusion

The Sigma Insight: Standard and General Equation of a Circle

Decoding the Circle Centers

Analyzing the Quadrilateral Sides

Verifying the Geometry

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the lines 2x+3y+1=0 and 3x−y−4=0 lie along diameter of a circle of circumference 10π, then the equation of the circle is

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Analyze Circle $M$

Analyze Circle $N$

Analyze Circle $O$

Analyze Circle $P$

If $m_{i}, \frac{1}{m _{i}}, m_{i} > 0, i = 1, 2, 3, 4$ are four distinct points on a circle, then show that $m_{1} m_{2} m_{3} m_{4} = 1$ .

If the lines $2 x + 3 y + 1 = 0$ and $3 x - y - 4 = 0$ lie along diameter of a circle of circumference $10 π$ , then the equation of the circle 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 :

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

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