Introduction to Circles

Q: What is the equation of a circle in JEE Main and Advanced problems?

The most useful forms are the standard form (x - h)^2 + (y - k)^2 = r^2 and the general form x^2 + y^2 + 2gx + 2fy + c = 0. The general form is preferred when the centre and radius are unknown, because you can substitute a point directly and set up linear equations in g, f, c.

Q: How do you find the centre and radius from the general equation of a circle?

From x^2 + y^2 + 2gx + 2fy + c = 0: centre is (-g,\, -f) and radius is \sqrt{g^2 + f^2 - c}. Read the coefficients of x and y, divide by 2 to get g and f, then apply the formula.

Q: When does a second-degree equation in x and y represent a circle?

When (i) coefficient of x^2 equals coefficient of y^2 and (ii) coefficient of xy is zero. If both are true, divide through to make the x^2 coefficient 1 and read off centre and radius as usual.

Q: What is the diameter form of a circle and why does it work?

If (x_1, y_1) and (x_2, y_2) are endpoints of a diameter, the circle is (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0. It works because the angle in a semicircle is 90^\circ, so any point P on the circle sees the diameter at a right angle, giving the perpendicular slope condition.

Q: How do you find the equation of a circle passing through three given points?

Substitute each of the three points into x^2 + y^2 + 2gx + 2fy + c = 0 to get three linear equations in g, f, c. Solve them and substitute back. The three points must be non-collinear.

Q: How do you find the length of the intercept a circle makes on the x-axis?

For x^2 + y^2 + 2gx + 2fy + c = 0, the x-intercept has length 2\sqrt{g^2 - c}. If g^2 \gt c the circle cuts the axis at two points, if g^2 = c it touches the axis, and if g^2 \lt c the circle does not meet the axis.

Q: What are the parametric equations of a circle?

For (x - h)^2 + (y - k)^2 = r^2: x = h + r\cos\theta,\; y = k + r\sin\theta, with \theta \in (-\pi, \pi]. The parameter \theta is the angle the radius through the point makes with the positive x-direction.

Q: How do you check whether a point lies inside or outside a circle?

Compute S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c at the point (x_1, y_1). If S_1 \lt 0 the point is inside, S_1 = 0 means on the circle, and S_1 \gt 0 means outside.

Q: What is a point circle and does it appear in JEE problems?

When g^2 + f^2 - c = 0, the equation x^2 + y^2 + 2gx + 2fy + c = 0 has radius 0 and represents just the single point (-g, -f). It shows up in JEE Main / Advanced as an edge case when locus problems collapse to a point.

MathsCirclesFor JEE aspirants

A circle is the locus of a point in a plane that moves at a constant distance (the radius) from a fixed point (the centre). The most common form used in JEE problems is the general equation $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , which has centre $(- g, - f)$ and radius $g^{2} + f^{2} - c$ . This concept introduces every form of the circle equation you need - standard, general, diameter and parametric - along with the position of a point and the intercepts a circle makes on the coordinate axes.

Key Formulas - Quick Reference

Circle with centre $(h, k)$ and radius $r$ : $(x - h)^{2} + (y - k)^{2} = r^{2}$
Circle centred at origin, radius $r$ : $x^{2} + y^{2} = r^{2}$
General equation: $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ - centre $(- g, - f)$ , radius $g^{2} + f^{2} - c$
Diameter form (endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ ): $(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$
Intercepts on axes for $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ : $2 g^{2} - c$ on x-axis, $2 f^{2} - c$ on y-axis
Parametric form of $(x - h)^{2} + (y - k)^{2} = r^{2}$ : $x = h + r cos θ, y = k + r sin θ, - π < θ \leq π$
Position of $(x_{1}, y_{1})$ w.r.t. $S \equiv x^{2} + y^{2} + 2 g x + 2 f y + c$ : inside if $S_{1} < 0$ , on if $S_{1} = 0$ , outside if $S_{1} > 0$

1. Definition of a Circle

A circle is the set of all points in a plane that are at a fixed distance $r$ from a fixed point $C$ . The fixed point $C$ is called the centre and the fixed distance $r$ is called the radius.

Every point

P (x, y)

on the circle satisfies

C P = r

, and this single condition gives us the equation of the circle in any form we like.

2. Equation of a Circle - Various Forms

(a) Standard form - centre at origin

A circle centred at the origin with radius $r$ has the equation

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

(b) Standard form - centre at $(h, k)$

A circle with centre $(h, k)$ and radius $r$ has the equation

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

Figure 1: Circle with centre (h, k) and radius r. Every point P(x, y) on it satisfies (x - h)² + (y - k)² = r².

(c) General equation of a circle

The general second-degree equation representing a circle is

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

with centre $(- g, - f)$ and radius $g^{2} + f^{2} - c$ .

Derivation. Expanding $(x - h)^{2} + (y - k)^{2} = r^{2}$ gives

x^{2} + y^{2} - 2 h x - 2 k y + (h^{2} + k^{2} - r^{2}) = 0.

Comparing with the general form, we take $- h = g$ , $- k = f$ and $h^{2} + k^{2} - r^{2} = c$ . Hence centre $= (- g, - f)$ and $r = g^{2} + f^{2} - c$ .

Condition to define a real circle:

$g^{2} + f^{2} - c > 0$ - real circle
$g^{2} + f^{2} - c = 0$ - point circle (a single point at $(- g, - f)$ )
$g^{2} + f^{2} - c < 0$ - imaginary circle (no real points; the "centre" $(- g, - f)$ still exists)

Recognising a circle from a second-degree equation. A general second-degree equation

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

represents a circle if and only if

coefficient of $x^{2}$ = coefficient of $y^{2}$ (i.e. $a = b$ ), and
coefficient of $x y$ is zero (i.e. $h = 0$ ).

Divide throughout by

a

to bring it to standard general form.

(d) Diameter form

If $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are the endpoints of a diameter of a circle, then the equation of the circle is

(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0.

Figure 2: Circle drawn on AB as diameter. For every point P(x, y) on the circle ∠APB = 90°, which gives (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0.

Reason: The angle in a semicircle is a right angle. So for any point $P (x, y)$ on the circle, $P A ⊥ P B$ , i.e. $(slope of P A) (slope of P B) = - 1$ . This gives

\frac{y - y _{1}}{x - x _{1}} \cdot \frac{y - y _{2}}{x - x _{2}} = - 1,

which simplifies to the diameter form. This is also the circle of least radius passing through the two given points.

(e) Circle through three given points

Three non-collinear points determine a unique circle. The standard route is:

Write the circle in general form $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ .
Substitute each of the three points in turn. This gives three linear equations in $g$ , $f$ and $c$ .
Solve the system for $g$ , $f$ , $c$ and put the values back into the general form.

Solved Example 1

Find the equation of the circle whose centre is

(1, - 2)

and radius is

4

Solution:

Using $(x - h)^{2} + (y - k)^{2} = r^{2}$ with $(h, k) = (1, - 2)$ and $r = 4$ :

$(x - 1)^{2} + (y + 2)^{2} = 16$

Expanding: $x^{2} + y^{2} - 2 x + 4 y - 11 = 0.$

Solved Example 2

Find the equation of the circle which passes through the point of intersection of the lines

3 x - 2 y - 1 = 0

and

4 x + y - 27 = 0

and whose centre is

(2, - 3)

Solution:

Solving $3 x - 2 y - 1 = 0$ and $4 x + y - 27 = 0$ simultaneously: $x = 5$ , $y = 7$ . So $P = (5, 7)$ .

Radius $= C P = (5 - 2)^{2} + (7 + 3)^{2} = 9 + 100 = 109$ .

Required circle: $(x - 2)^{2} + (y + 3)^{2} = 109.$

Solved Example 3

Find the centre and radius of the circle

x^{2} + y^{2} - 4 x + 6 y + 12 = 0

Solution:

Comparing with $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ :

$2 g = - 4 \Rightarrow g = - 2$ ; $2 f = 6 \Rightarrow f = 3$ ; $c = 12$ .

Centre $= (- g, - f) = (2, - 3) .$

Radius $= g^{2} + f^{2} - c = 4 + 9 - 12 = 1.$

Solved Example 4

Find the equation of the circle whose diameter has endpoints

(- 1, 2)

and

(4, - 3)

Solution:

Using diameter form with $x_{1} = - 1, x_{2} = 4, y_{1} = 2, y_{2} = - 3$ :

$(x + 1) (x - 4) + (y - 2) (y + 3) = 0$

$\Rightarrow x^{2} + y^{2} - 3 x + y - 10 = 0.$

Solved Example 5

Find the equation of the circle passing through the three points

(1, 1)

(2, - 1)

and

(3, 2)

Solution:

Let the circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . Substituting each point:

$(1, 1)$ : $2 + 2 g + 2 f + c = 0 \Rightarrow 2 g + 2 f + c = - 2$ ...(i)

$(2, - 1)$ : $5 + 4 g - 2 f + c = 0 \Rightarrow 4 g - 2 f + c = - 5$ ...(ii)

$(3, 2)$ : $13 + 6 g + 4 f + c = 0 \Rightarrow 6 g + 4 f + c = - 13$ ...(iii)

Solving (i), (ii), (iii): $g = - \frac{5}{2}, f = - \frac{1}{2}, c = 4$ .

Required circle: $x^{2} + y^{2} - 5 x - y + 4 = 0.$

3. Intercepts Made by a Circle on the Axes

For the circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , the length of the chord it cuts off on the x-axis is $2 g^{2} - c$ and on the y-axis is $2 f^{2} - c$ .

Figure 3: The circle cuts the x-axis at A and B. Dropping a perpendicular from C(-g, -f) to the axis at D gives CD = |f|, so AB = 2 · AD = 2√(g² - c).

Nature of x-intercept depends on the sign of $g^{2} - c$ :

Condition	What it means
$g^{2} > c$	Circle cuts the x-axis at two distinct points
$g^{2} = c$	Circle touches the x-axis (tangent from inside)
$g^{2} < c$	Circle lies entirely above or below the x-axis (no intercept)

The same three cases apply to the y-axis with $f^{2}$ in place of $g^{2}$ .

Derivation. Drop a perpendicular from the centre $C (- g, - f)$ to the x-axis, meeting it at $D$ . Since a perpendicular from the centre bisects a chord, $D$ is the midpoint of $A B$ and $C D = ∣ f ∣$ . In right triangle $C D A$ ,

A D = r^{2} - C D^{2} = (g^{2} + f^{2} - c) - f^{2} = g^{2} - c,

so the intercept is $A B = 2 A D = 2 g^{2} - c$ . Replacing $f$ by $g$ gives the y-axis intercept $2 f^{2} - c$ .

Solved Example 6

Find the equation of the circle touching the negative y-axis at a distance

3

from the origin and cutting an intercept of length

8

on the x-axis.

Solution:

Let the circle be $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . It touches the y-axis, so $f^{2} = c$ ...(i).

The point $(0, - 3)$ lies on it, so $9 - 6 f + c = 0$ ...(ii).

From (i) and (ii): $9 - 6 f + f^{2} = 0 \Rightarrow (f - 3)^{2} = 0 \Rightarrow f = 3$ , and then $c = 9$ .

x-intercept: $2 g^{2} - c = 8 \Rightarrow g^{2} - 9 = 16 \Rightarrow g = \pm 5$ .

Required circle: $x^{2} + y^{2} \pm 10 x + 6 y + 9 = 0.$

4. Parametric Equations of a Circle

The parametric equations of the circle $(x - h)^{2} + (y - k)^{2} = r^{2}$ are

x = h + r cos θ, y = k + r sin θ, - π < θ \leq π,

where $(h, k)$ is the centre, $r$ is the radius and $θ$ is the parameter (the angle the radius through $P$ makes with the positive x-direction).

Figure 4: Parametric form. Every point of (x - h)² + (y - k)² = r² can be written as P(h + r cos θ, k + r sin θ), where θ is measured from the horizontal through the centre.

Solved Example 7

Find the parametric equations of the circle

x^{2} + y^{2} - 4 x - 2 y + 1 = 0

Solution:

Complete the square: $(x^{2} - 4 x) + (y^{2} - 2 y) = - 1 \Rightarrow (x - 2)^{2} + (y - 1)^{2} = 4 = 2^{2} .$

Centre $(2, 1)$ , radius $2$ . Parametric form:

$x = 2 + 2 cos θ, y = 1 + 2 sin θ .$

Solved Example 8

Convert the parametric curve

x = a + c cos θ, y = b + c sin θ

to Cartesian form and identify centre and radius.

Solution:

From the given equations, $cos θ = \frac{x - a}{c}$ and $sin θ = \frac{y - b}{c} .$

Using $cos^{2} θ + sin^{2} θ = 1$ :

$(\frac{x - a}{c})^{2} + (\frac{y - b}{c})^{2} = 1 \Rightarrow (x - a)^{2} + (y - b)^{2} = c^{2} .$

Circle with centre $(a, b)$ and radius $c$ .

5. Position of a Point with Respect to a Circle

Let $S \equiv x^{2} + y^{2} + 2 g x + 2 f y + c$ . For any point $(x_{1}, y_{1})$ , define

S_{1} \equiv x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c .

Then the point $(x_{1}, y_{1})$ lies

inside the circle if $S_{1} < 0$ ,
on the circle if $S_{1} = 0$ ,
outside the circle if $S_{1} > 0$ .

Figure 5: Comparing CP with the radius r. A point lies inside when S₁ < 0, on the circle when S₁ = 0 and outside when S₁ > 0.

Greatest and least distances. If a point

A

lies outside a circle with centre

C

and radius

r

, then the greatest and least distances from

A

to the circle are

A C + r

and

A C - r

respectively (along the line through

A

and

C

Solved Example 9

Discuss the position of the points

(1, 2)

and

(6, 0)

with respect to the circle

x^{2} + y^{2} - 4 x + 2 y - 11 = 0

Solution:

Let $S \equiv x^{2} + y^{2} - 4 x + 2 y - 11$ .

At $(1, 2)$ : $S_{1} = 1 + 4 - 4 + 4 - 11 = - 6 < 0$ , so $(1, 2)$ lies inside.

At $(6, 0)$ : $S_{2} = 36 + 0 - 24 + 0 - 11 = 1 > 0$ , so $(6, 0)$ lies outside.

Solved Example 10

How are the points

(0, 1)

(3, 1)

and

(1, 3)

situated with respect to the circle

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

Solution:

Let $S \equiv x^{2} + y^{2} - 2 x - 4 y + 3$ .

At $(0, 1)$ : $0 + 1 - 0 - 4 + 3 = 0$ , so $(0, 1)$ lies on the circle.

At $(3, 1)$ : $9 + 1 - 6 - 4 + 3 = 3 > 0$ , so $(3, 1)$ lies outside.

At $(1, 3)$ : $1 + 9 - 2 - 12 + 3 = - 1 < 0$ , so $(1, 3)$ lies inside.

Common Mistakes to Avoid

Watch out

Sign of $g$ and $f$ when reading off the centre. From $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ the centre is $(- g, - f)$ , not $(g, f)$ . A coefficient like $- 4 x$ means $2 g = - 4$ , so $g = - 2$ and the x-coordinate of the centre is $+ 2$ .
Forgetting the "coefficient of $x^{2}$ = coefficient of $y^{2}$ " check. Before you claim an equation is a circle, divide through so both coefficients are $1$ , and verify there is no $x y$ term.
Confusing "point circle" with "no circle". When $g^{2} + f^{2} - c = 0$ the equation represents the single point $(- g, - f)$ , not "no solution".
Forgetting the diameter form uses two endpoints of a diameter, not just any two points on the circle. A general chord's endpoints do not satisfy this form.
Dropping the modulus in intercept length. The x-intercept length is $2 g^{2} - c$ ; it is a length, so take the non-negative root only.
Restricting the parameter $θ$ unnecessarily. $θ \in (- π, π]$ covers the full circle exactly once; students sometimes restrict to $[0, 2 π)$ instead - both are fine, but pick one.

Frequently Asked Questions

What is the equation of a circle in JEE Main and Advanced problems?

The most useful forms are the standard form $(x - h)^{2} + (y - k)^{2} = r^{2}$ and the general form $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ . The general form is preferred when the centre and radius are unknown, because you can substitute a point directly and set up linear equations in $g$ , $f$ , $c$ .

How do you find the centre and radius from the general equation of a circle?

From $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ : centre is $(- g, - f)$ and radius is $g^{2} + f^{2} - c$ . Read the coefficients of $x$ and $y$ , divide by $2$ to get $g$ and $f$ , then apply the formula.

When does a second-degree equation in x and y represent a circle?

When (i) coefficient of $x^{2}$ equals coefficient of $y^{2}$ and (ii) coefficient of $x y$ is zero. If both are true, divide through to make the $x^{2}$ coefficient $1$ and read off centre and radius as usual.

What is the diameter form of a circle and why does it work?

If $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are endpoints of a diameter, the circle is $(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$ . It works because the angle in a semicircle is $9 0^{\circ}$ , so any point $P$ on the circle sees the diameter at a right angle, giving the perpendicular slope condition.

How do you find the equation of a circle passing through three given points?

Substitute each of the three points into $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ to get three linear equations in $g$ , $f$ , $c$ . Solve them and substitute back. The three points must be non-collinear.

How do you find the length of the intercept a circle makes on the x-axis?

For $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ , the x-intercept has length $2 g^{2} - c$ . If $g^{2} > c$ the circle cuts the axis at two points, if $g^{2} = c$ it touches the axis, and if $g^{2} < c$ the circle does not meet the axis.

What are the parametric equations of a circle?

For $(x - h)^{2} + (y - k)^{2} = r^{2}$ : $x = h + r cos θ, y = k + r sin θ$ , with $θ \in (- π, π]$ . The parameter $θ$ is the angle the radius through the point makes with the positive x-direction.

How do you check whether a point lies inside or outside a circle?

Compute $S_{1} = x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c$ at the point $(x_{1}, y_{1})$ . If $S_{1} < 0$ the point is inside, $S_{1} = 0$ means on the circle, and $S_{1} > 0$ means outside.

What is a point circle and does it appear in JEE problems?

When $g^{2} + f^{2} - c = 0$ , the equation $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ has radius $0$ and represents just the single point $(- g, - f)$ . It shows up in JEE Main / Advanced as an edge case when locus problems collapse to a point.

Previous year questions on Introduction to Circles

6 questions from past papers, each with a step-by-step solution.

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Practice questions →Take a mock test

1. Definition of a Circle

2. Equation of a Circle - Various Forms

(a) Standard form - centre at origin

(b) Standard form - centre at (h,k)

(c) General equation of a circle

(d) Diameter form

(e) Circle through three given points

3. Intercepts Made by a Circle on the Axes

4. Parametric Equations of a Circle

5. Position of a Point with Respect to a Circle

Common Mistakes to Avoid

Frequently Asked Questions

What is the equation of a circle in JEE Main and Advanced problems?

How do you find the centre and radius from the general equation of a circle?

When does a second-degree equation in x and y represent a circle?

What is the diameter form of a circle and why does it work?

How do you find the equation of a circle passing through three given points?

How do you find the length of the intercept a circle makes on the x-axis?

What are the parametric equations of a circle?

How do you check whether a point lies inside or outside a circle?

What is a point circle and does it appear in JEE problems?

Previous year questions on Introduction to Circles

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(b) Standard form - centre at $(h, k)$