The equation of a circle

What you will be able to do

• write the equation of a circle from its centre and radius;
• read the centre and radius from the general equation, and say when it is not a circle;
• find a circle through three points or on a given diameter;
• use the parametric form of a circle.

A sprinkler on a lawn

A sprinkler stands at a point on a lawn and wets everything within $5$ metres. The edge of the wet patch is a circle. Every point on that edge is $5$ metres from the sprinkler, and no other point is. That is all a circle is. In coordinates the sentence becomes an equation, and then geometry turns into algebra.

You already know the distance formula from the chapter on straight lines. You use it here for a point $P(x, y)$ and a fixed centre $C(h, k)$.

The standard form

The point $P(x, y)$ is on the circle with centre $C(h, k)$ and radius $r$ exactly when $CP = r$. By the distance formula,

Squaring both sides gives the standard form

If the centre is the origin, $h = k = 0$ and the equation is $x^2 + y^2 = r^2$. A-Level books write the same thing as $(x - a)^2 + (y - b)^2 = r^2$. Only the letters change.

The general form

Open the brackets of the standard form and collect the terms. You get

Turn it round. The circle with the general equation has

• centre $(-g, -f)$, which is minus half the coefficient of $x$ and minus half the coefficient of $y$;
• radius $r = \sqrt{g^2 + f^2 - c}$.

Many books write the same equation as $x^2 + y^2 + Dx + Ey + F = 0$, and then the centre is $\left(-\dfrac{D}{2}, -\dfrac{E}{2}\right)$. JEE questions usually use $g$, $f$ and $c$. The idea is the same.

Look at the number $g^2 + f^2 - c$ under the root.

• If it is positive, the equation is a circle.
• If it is zero, the radius is $0$ and the "circle" is the single point $(-g, -f)$.
• If it is negative, no real point satisfies the equation. Some books say the circle is imaginary.

When is a second degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ a circle? It needs $a = b$ and $h = 0$, which means the same coefficient on $x^2$ and $y^2$ and no $xy$ term. If the coefficients of $x^2$ and $y^2$ are both $2$, divide the whole equation by $2$ before you read $g$, $f$ and $c$.

A worked example

Write the circle with centre $(3, -2)$ and radius $5$ in both forms.

1. Standard form. $(x - 3)^2 + (y + 2)^2 = 25$.
2. Open the brackets. $x^2 - 6x + 9 + y^2 + 4y + 4 = 25$.
3. General form. $x^2 + y^2 - 6x + 4y - 12 = 0$. So $g = -3$, $f = 2$ and $c = -12$.
4. Check the radius. $\sqrt{g^2 + f^2 - c} = \sqrt{9 + 4 + 12} = 5$.

Now go the other way. Find the centre and radius of $x^2 + y^2 - 6x + 8y - 24 = 0$. Here $2g = -6$ and $2f = 8$, so $g = -3$, $f = 4$ and $c = -24$. The centre is $(3, -4)$. The radius is $\sqrt{9 + 16 + 24} = \sqrt{49} = 7$.

Three points fix a circle

A circle has three unknowns: $g$, $f$ and $c$ (or $h$, $k$ and $r$). So three points on it fix the circle, as long as they are not on one line. Put each point into $x^2 + y^2 + 2gx + 2fy + c = 0$. You get three linear equations in $g$, $f$, $c$.

Find the circle through $A(6, 0)$, $B(-1, 7)$ and $D(2, -2)$.

1. Three equations. $36 + 12g + c = 0$, $50 - 2g + 14f + c = 0$ and $8 + 4g - 4f + c = 0$.
2. Subtract. The first minus the third gives $28 + 8g + 4f = 0$, so $2g + f = -7$. The second minus the third gives $42 - 6g + 18f = 0$, so $-g + 3f = -7$.
3. Solve. From the second, $g = 3f + 7$. Then $2(3f + 7) + f = -7$, so $7f = -21$, $f = -3$ and $g = -2$. From the first equation $c = -36 - 12g = -12$.
4. The circle. $x^2 + y^2 - 4x - 6y - 12 = 0$. Its centre is $(2, 3)$ and its radius is $\sqrt{4 + 9 + 12} = 5$.

There is a geometric way too. The centre is the same distance from all three points, so it lies on the perpendicular bisector of each pair. Two bisectors meet at the centre, as the figure shows.

Circle on a diameter

A line segment $AB$ with $A(x_1, y_1)$ and $B(x_2, y_2)$ is a diameter of a circle exactly when the angle $APB$ is a right angle for every other point $P$ of the circle. That fact is the angle in a semicircle from Class 10. In coordinates it says that the slopes of $PA$ and $PB$ multiply to $-1$, which gives the diameter form

For $A(1, 2)$ and $B(7, 10)$ it gives $x^2 + y^2 - 8x - 12y + 27 = 0$. The centre is the midpoint $(4, 6)$ and the radius is $\sqrt{16 + 36 - 27} = 5$. This is half of $AB = 10$, as it must be.

Circles that touch the axes

The distance of the centre $(-g, -f)$ from the $y$-axis is $|g|$, and from the $x$-axis it is $|f|$. A circle touches the $x$-axis when $r = |f|$, which means $g^2 + f^2 - c = f^2$, so $c = g^2$. In the same way it touches the $y$-axis when $c = f^2$. A circle that touches both axes has $|g| = |f| = r$, so its centre is $(\pm r, \pm r)$.

For instance, the circle with centre $(4, -3)$ and radius $3$ touches the $x$-axis. Its equation is $x^2 + y^2 - 8x + 6y + 16 = 0$, and $c = 16 = g^2$.

The parametric form

You can name any point of the circle with one angle. For the circle with centre $(h, k)$ and radius $r$,

As $\theta$ goes from $0$ to $2\pi$, the point goes once round the circle. The entrance exams use this parametric form for locus questions, and A-Level courses meet it again in the topic of parametric equations.

Practise with new numbers

Try it in Python

The program finds the circle through three points by solving the three linear equations, and it checks that each point is on the circle.

Common mistakes

• Reading the centre with the wrong sign. In $(x + 2)^2 + (y - 5)^2 = 9$ the centre is $(-2, 5)$, not $(2, -5)$.
• Forgetting to divide. In $2x^2 + 2y^2 - 8x + 4y - 5 = 0$ divide by $2$ first. Otherwise $g$, $f$ and $c$ are wrong.
• Leaving out the square root. The radius is $\sqrt{g^2 + f^2 - c}$. The number under the root is $r^2$.
• Calling every second degree equation a circle. An $xy$ term, or different coefficients of $x^2$ and $y^2$, rule it out.

Remember these

• The circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.
• $x^2 + y^2 + 2gx + 2fy + c = 0$ has centre $(-g, -f)$ and radius $\sqrt{g^2 + f^2 - c}$.
• The circle on the ends $(x_1, y_1)$ and $(x_2, y_2)$ of a diameter is $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.
• A circle touches the $x$-axis when $c = g^2$ and the $y$-axis when $c = f^2$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Write the equation of the circle with centre $(-1, 4)$ and radius $3$ in general form.

2. Does $x^2 + y^2 + 4x - 6y + 20 = 0$ represent a circle?

3. Find the circle on the segment joining $(2, -1)$ and $(-4, 5)$ as a diameter. What is its radius?

4. For what value of $k$ is $x^2 + y^2 - 6x + 2y + k = 0$ a circle of radius $4$?

5. Find the point at $\theta = 60^\circ$ on the circle $x^2 + y^2 - 4x + 2y - 4 = 0$.

References

• NCERT, Mathematics for Class XI, Chapter 10 (Conic Sections), the section on the circle.
• OpenStax, Precalculus 2e, Chapter 12 (Analytic Geometry) and Chapter 2 on the distance formula and the equation of a circle.