Circle

What is a circle?

In plain words: a circle is every point at one fixed distance r from one fixed point C. It is the section of a cone made by a plane perpendicular to the axis and not through the vertex (NCERT Section 10.2.1); through the vertex, that plane cuts only a point (NCERT Section 10.2.2). With coordinates, the definition becomes an equation.

The standard equation

Let the centre be C(h, k) and the radius r, and let P(x, y) be any point of the circle (NCERT page 180, Fig 10.12). By the definition, CP = r. The distance formula turns this into √((x − h)² + (y − k)²) = r. Both sides are positive, so squaring loses nothing:

The equation of the circle is simplest if the centre of the circle is at the origin.

NCERT Mathematics (Class 11), Section 10.3, page 180

With the centre at the origin, h = k = 0 and the circle is x² + y² = r² (NCERT Example 1). NCERT Example 2 (page 180): centre (−3, 2) and radius 4 give (x + 3)² + (y − 2)² = 16. In the Argand plane, the same circle is the set of complex numbers z with |z − z₀| = r (see modulus as a distance).

Centre and radius from the general equation

Multiplying out (x − h)² + (y − k)² = r² gives x² + y² − 2hx − 2ky + (h² + k² − r²) = 0. So a circle’s equation has equal coefficients of x² and y² and no xy term. To read off the centre and radius, collect the x terms and the y terms and complete the square in each: add (half the coefficient of x)² and (half the coefficient of y)² to both sides, the step behind the quadratic formula.

NCERT Example 3 (page 180): x² + y² + 8x + 10y − 8 = 0.

(x² + 8x) + (y² + 10y) = 8

Add (8/2)² = 16 and (10/2)² = 25 to both sides:

(x² + 8x + 16) + (y² + 10y + 25) = 8 + 16 + 25

(x + 4)² + (y + 5)² = 49

So the centre is (−4, −5) and the radius is 7.

The general form (JEE). Doing the same with x² + y² + 2gx + 2fy + c = 0 gives (x + g)² + (y + f)² = g² + f² − c:

If g² + f² − c = 0, the “circle” is just the point (−g, −f), a point circle. If it is negative, no real point satisfies the equation (an imaginary circle). The coefficients of x² and y² must be 1 before you read off g and f: divide 2x² + 2y² − x = 0 by 2 first.

A circle from given conditions

A circle has three unknowns, h, k and r, so it needs three independent conditions. When one condition is quadratic, there can be two circles (see the solved Example 4 below, NCERT Exercise 10.1, Q12). NCERT Example 4 (page 181) asks for the circle through (2, −2) and (3, 4) whose centre is on x + y = 2. The centre is equally far from the two points:

(2 − h)² + (−2 − k)² = (3 − h)² + (4 − k)²

The h² and k² terms cancel:

8 − 4h + 4k = 25 − 6h − 8k

2h + 12k = 17.

With h + k = 2 this gives k = 1.3 and h = 0.7. Then r² = 1.3² + 3.3² = 12.58, and the circle is (x − 0.7)² + (y − 1.3)² = 12.58. NCERT gives only the final values.

Inside, on or outside (JEE). A point P is inside, on or outside the circle as CP is less than, equal to or greater than r. In the form S = (x − h)² + (y − k)² − r², put in the point and look at the sign of S: negative inside, zero on, positive outside. The same works straight from the general form: x² + y² + 2gx + 2fy + c equals (x + g)² + (y + f)² − r², so put the point into the left side (after making the coefficient of x² equal to 1) and read the sign.

Solved examples

Common mistakes to avoid

Quick revision summary

• A circle is the set of points at a fixed distance r (the radius) from a fixed point C (the centre).
• Centre (h, k), radius r: (x − h)² + (y − k)² = r²; centre at the origin: x² + y² = r².
• x² + y² + 2gx + 2fy + c = 0 has centre (−g, −f) and radius √(g² + f² − c); complete the squares to see it.
• A circle needs three conditions; a point is inside, on or outside as its distance from the centre is less than, equal to or more than r.

Questions students ask

What is the difference between the standard and general equation of a circle?

The standard equation of a circle, (x − h)² + (y − k)² = r², shows the centre (h, k) and the radius r at a glance, while the general equation x² + y² + 2gx + 2fy + c = 0 is the same circle multiplied out, with its centre at (−g, −f). Every circle has both forms, and you pass between them by expanding or by completing the squares; but an equation of the general shape is a real circle only when g² + f² − c, the radius squared, is positive. For example, (x − 1)² + (y + 3)² = 9 expands to x² + y² − 2x + 6y + 1 = 0.

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

To find the equation of a circle through three points, put each point into x² + y² + 2gx + 2fy + c = 0 and solve the three linear equations for g, f and c. Exactly one circle passes through three points that are not on a line, and its centre is the circumcentre of their triangle; three distinct collinear points lie on no circle, and two points lie on infinitely many. For example, (6, 2), (1, 7) and (−4, 2) give g = −1, f = −2 and c = −20, so the circle is x² + y² − 2x − 4y − 20 = 0, with centre (1, 2) and radius 5.

What does c mean in the general equation of a circle?

In the general equation of a circle x² + y² + 2gx + 2fy + c = 0, the constant c is the value of the left side at the origin, so by the inside or outside test its sign tells where the origin lies, for a real circle: inside it if c < 0, on it if c = 0 and outside it if c > 0. In terms of the centre and radius, c = h² + k² − r², the square of the centre’s distance from the origin minus the square of the radius. For example, x² + y² − 6x − 2y + 6 = 0 has c = 6 > 0: its centre (3, 1) is √10 from the origin, more than its radius 2.

What is the equation of the unit circle?

The unit circle is the circle of radius 1 centred at the origin, so its equation is x² + y² = 1. Each of its points is (cos θ, sin θ) for some angle θ, which is how the trigonometric functions are defined, and this is why cos² θ + sin² θ = 1. In the same way, every point of the circle with centre (h, k) and radius r is (h + r cos θ, k + r sin θ), the parametric form of a circle, beyond NCERT.

Is circle a separate chapter in Class 11 maths?

The circle is not a separate chapter in the NCERT Class 11 Mathematics book (2026–27 reprint): it is Section 10.3 of Chapter 10, Conic Sections, which derives the standard equation and finds the centre and radius by completing the square. The JEE Main syllabus (Unit 10) also names the general form, which NCERT uses only in examples, and two topics the NCERT book leaves out, the circle on given ends of a diameter and the points where a line meets a circle centred at the origin.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Find the equation of the circle with centre (0, 2) and radius 2 (NCERT Exercise 10.1, Q1).

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   x² + (y − 2)² = 4, that is x² + y² − 4y = 0.

2. Q2. Find the equation of the circle with centre (−2, 3) and radius 4 (NCERT Exercise 10.1, Q2).

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   (x + 2)² + (y − 3)² = 16, that is x² + y² + 4x − 6y − 3 = 0.

3. Q3. Find the centre and radius of (x + 5)² + (y − 3)² = 36 (NCERT Exercise 10.1, Q6).

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   Centre (−5, 3), radius 6.

4. Q4. Find the centre and radius of x² + y² − 8x + 10y − 12 = 0 (NCERT Exercise 10.1, Q8).

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   (x − 4)² + (y + 5)² = 12 + 16 + 25 = 53, so the centre is (4, −5) and the radius √53.