MathsISC Class 11 · Circles

The Circle: Standard Form, General Form, Diameter Form and Parametric Form

Q: Why must the coefficients of x² and y² be equal?

In a circle every direction is equivalent, so x² and y² must have the same coefficient and there is no xy term.

Q: What is a point circle?

A circle of radius 0, a single point.

Q: How do I find the diameter?

Twice the radius: 2\sqrt{g^2 + f^2 - c}.

Updated 17 October 2026 · Fact-checked

A circle with centre (h, k) and radius r has the equation (x − h)² + (y − k)² = r². The general form x² + y² + 2gx + 2fy + c = 0 has centre (−g, −f) and radius √(g² + f² − c). A circle with the end points of a diameter (x₁, y₁) and (x₂, y₂) is (x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0.

Every point P on the circle sees diameter AB at a right angle, and CP stays equal to r, which gives the three equation forms.

Illustration drawn for QuestionBuddy.

Understand Equation of a Circle: Standard, General and Diameter Forms

Definition. A circle is the locus of a point at a fixed distance r (the radius) from a fixed point (the centre). Using the distance formula gives $(x - h)^{2} + (y - k)^{2} = r^{2}$ . With the centre at the origin: $x^{2} + y^{2} = r^{2}$ .

General form. Expanding the standard form: $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ with $g = - h$ , $f = - k$ , $c = h^{2} + k^{2} - r^{2}$ . So the centre is $(- g, - f)$ and the radius is $r = g^{2} + f^{2} - c$ . The equation represents a real circle if $g^{2} + f^{2} - c > 0$ , a point circle if it is 0 and no real circle if negative. To find the centre and radius from a general equation with coefficient of $x^{2}$ not 1, divide through by that coefficient first.

Diameter form. If $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ are the ends of a diameter, then for every point P(x, y) on the circle angle APB is a right angle, giving $(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$ .

Parametric form. $x = h + r cos θ$ , $y = k + r sin θ$ .

Position of a point. For the circle $S = x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ and a point $P (x_{1}, y_{1})$ : if $S_{1} = x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c$ is negative, P is inside; if zero, on the circle; if positive, outside.

Finding a circle. Use the standard form if the centre and radius are given; use the general form if three conditions on g, f, c are given (points on the circle).

Equation of a Circle: Standard, General and Diameter Forms: formulas to remember

Standard form: $$(x - h)^2 + (y - k)^2 = r^2$$; Centre (h, k).
General form: $x^2 + y^2 + 2gx + 2fy + c = 0$, centre $(-g, -f)$, $r = \sqrt{g^2 + f^2 - c}$; Real circle if $g^{2} + f^{2} > c$ .
Diameter form: $$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$; From the right angle in a semicircle.
Point position: $$S_1 < 0$ inside | $S_1 = 0$ on | $S_1 > 0$ outside$; S₁ = value of S at the point.

Equation of a Circle: Standard, General and Diameter Forms in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A circular fountain

Suppose a circular fountain has centre (2, −3) and radius 5 m on a plan.

Write its equation in standard and general form and decide whether a lamp at (6, 0) is inside or outside.

Try it first, then show the solutionSolution

Standard: $(x - 2)^{2} + (y + 3)^{2} = 25$ .
Expand: $x^{2} + y^{2} - 4 x + 6 y + 4 + 9 - 25 = 0$ , i.e. $x^{2} + y^{2} - 4 x + 6 y - 12 = 0$ .
At (6, 0): $36 + 0 - 24 + 0 - 12 = 0$ , so the lamp is on the circle.

Answer: (x−2)² + (y+3)² = 25; the lamp is exactly on the circle.

Memory hook: Substitute the point into the general form.

Tricks to remember Equation of a Circle: Standard, General and Diameter Forms

Short memory aids. Say each one aloud once, then use it on a question the same day.

Half the coefficients
The centre is minus half of the coefficients of x and y.
Use it when: Reading the centre.
Divide first
For 2x² + 2y² + …, divide by 2 before reading g, f, c.
Use it when: Equations not in standard form.

Equation of a Circle: Standard, General and Diameter Forms: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Centre of a circle · 1 mark · MCQBoard level

The centre of the circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ is:
(a) $(3, - 2)$
(b) $(- 3, 2)$
(c) $(6, - 4)$
(d) $(3, 2)$

Try it first, then show the solutionSolution

$g = - 3$ , $f = 2$ , so the centre is $(- g, - f) = (3, - 2)$ .

Answer: (a) $(3, - 2)$

Example 2: Equation of a circle · 3 marks · Short answerBoard level

A circular pond has its centre at (−3, 4) and radius 6 on a plan. Write its equation in standard and general forms.

Try it first, then show the solutionSolution

Standard: $(x + 3)^{2} + (y - 4)^{2} = 36$ . (1 mark)
Expand: $x^{2} + 6 x + 9 + y^{2} - 8 y + 16 = 36$ . (1 mark)
General: $x^{2} + y^{2} + 6 x - 8 y - 11 = 0$ . (1 mark)

Answer: x² + y² + 6x − 8y − 11 = 0

Example 3: Radius from the general form · 4 marks · Short answerHarder board level

Find the centre and radius of $2 x^{2} + 2 y^{2} - 8 x + 12 y + 10 = 0$ . Is the point (1, 1) inside, on or outside the circle?

Try it first, then show the solutionSolution

Divide by 2: $x^{2} + y^{2} - 4 x + 6 y + 5 = 0$ . (1 mark)
$g = - 2$ , $f = 3$ , $c = 5$ : centre $(2, - 3)$ , radius $4 + 9 - 5 = 8 = 22$ . (1.5 marks)
At (1, 1): $1 + 1 - 4 + 6 + 5 = 9 > 0$ , so the point is outside. (1.5 marks)

Answer: Centre (2, −3); radius 2√2; (1, 1) is outside.

Example 4: Circle on a diameter · 4 marks · Short answerHarder board level

The end points of a diameter of a circle are (1, 2) and (5, 6). Find its equation, centre and radius.

Try it first, then show the solutionSolution

$(x - 1) (x - 5) + (y - 2) (y - 6) = 0$ . (1 mark)
$x^{2} - 6 x + 5 + y^{2} - 8 y + 12 = 0$ , i.e. $x^{2} + y^{2} - 6 x - 8 y + 17 = 0$ . (1.5 marks)
Centre $(3, 4)$ (the midpoint); $r^{2} = 9 + 16 - 17 = 8$ , $r = 22$ . (1 mark)
Check: half of $16 + 16 = 42$ is $22$ . (0.5 mark)

Answer: x² + y² − 6x − 8y + 17 = 0; centre (3, 4); r = 2√2.

Example 5: Finding k and a parameter · 5 marks · Long answerHOTS / competency

Find k such that $x^{2} + y^{2} - 4 x - 6 y + k = 0$ is a circle of radius 3. For which values of k is the equation a real circle? Write the parametric form of the circle of radius 3 with centre (2, 3).

Try it first, then show the solutionSolution

$g = - 2$ , $f = - 3$ : $r^{2} = 4 + 9 - k = 9$ , so $k = 4$ . (1.5 marks)
A real circle needs $13 - k > 0$ , so $k < 13$ (k = 13 is a point circle). (1.5 marks)
Parametric form: $x = 2 + 3 cos θ$ , $y = 3 + 3 sin θ$ . (1.5 marks)
Check: $(x - 2)^{2} + (y - 3)^{2} = 9 (cos^{2} θ + sin^{2} θ) = 9$ . (0.5 mark)

Answer: k = 4; k < 13; x = 2 + 3cosθ, y = 3 + 3sinθ.

Where students lose marks in Equation of a Circle: Standard, General and Diameter Forms

The slips examiners see most often, and what to write instead.

Mistake: Reading the centre with the wrong sign
Writing (−3, 2) for x² + y² − 6x + 4y.
Do this: The centre is (−g, −f) = (3, −2).
Usually costs: 2 marks.
Mistake: Not dividing by the coefficient of x²
Using g and f from 2x² + 2y² + …
Do this: Divide by 2 first.
Usually costs: 2 marks.
Mistake: Forgetting r²
Putting r on the right side.
Do this: The standard form has r².
Usually costs: 1 mark.

Exam tips

Compare with the general form term by term.
Check that g² + f² − c > 0.
Use the midpoint for the centre of a diameter.
Check the answer by substituting the given points.

Practice questions on Equation of a Circle: Standard, General and Diameter Forms

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The equation of the circle with centre at the origin and radius 4 is:
(a) $x^{2} + y^{2} = 4$
(b) $x^{2} + y^{2} = 16$
(c) $x^{2} + y^{2} = 8$
(d) $x + y = 4$
Hint
r² on the right.
Answer
(b) $x^{2} + y^{2} = 16$
Q2 · 2 marks · Short answerBoard level
Find the radius of $x^{2} + y^{2} + 6 x - 8 y = 0$ .
Hint
c = 0.
Answer
$g = 3$ , $f = - 4$ , $c = 0$ : $r = 9 + 16 = 5$ .
Q3 · 3 marks · Short answerHarder board level
Find the equation of the circle with centre (−1, 2) that passes through the point (3, −1).
Hint
Distance to the point gives r.
Answer
$r^{2} = 16 + 9 = 25$ ; $(x + 1)^{2} + (y - 2)^{2} = 25$ .
Q4 · 3 marks · Short answerHarder board level
Does $x^{2} + y^{2} - 2 x + 4 y + 10 = 0$ represent a real circle?
Hint
Check g² + f² − c.
Answer
$g = - 1$ , $f = 2$ , $c = 10$ : $1 + 4 - 10 = - 5 < 0$ , so there is no real circle.
Q5 · 3 marks · Short answerHarder board level
Find the equation of the circle having (0, 0) and (6, 8) as the ends of a diameter.
Hint
Diameter form.
Answer
$x (x - 6) + y (y - 8) = 0$ , i.e. $x^{2} + y^{2} - 6 x - 8 y = 0$ ; centre (3, 4), radius 5.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a radar station at (4, 3) detects aircraft within 10 km (the plan uses km).
(i) Write the equation of the detection circle. (1 mark)
(ii) Is an aircraft at (10, 11) detected? (1 mark)
(iii) Is an aircraft at (12, 9) detected? (1 mark)
(iv) Find the greatest detection distance from the origin. (1 mark)
Hint
Compare the distance with 10.
Answer
(i) $(x - 4)^{2} + (y - 3)^{2} = 100$ . (ii) Distance $= 36 + 64 = 10$ : on the boundary, detected just at the limit. (iii) Distance $= 64 + 36 = 10$ : also on the boundary. (iv) $5 + 10 = 15$ km (distance of the centre from the origin is 5).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Centre of x² + y² + 2gx + 2fy + c = 0?Show answer
(−g, −f).
2.Radius?Show answer
√(g² + f² − c).
3.Diameter form?Show answer
(x−x₁)(x−x₂) + (y−y₁)(y−y₂) = 0.
4.Point circle?Show answer
g² + f² = c.

Equation of a Circle: Standard, General and Diameter Forms: frequently asked questions

Why must the coefficients of x² and y² be equal?

In a circle every direction is equivalent, so x² and y² must have the same coefficient and there is no xy term.

What is a point circle?

A circle of radius 0, a single point.

How do I find the diameter?

Twice the radius: 2\sqrt{g^2 + f^2 - c}.