What Is the Equation of a Circle?
The equation of a circle is the algebraic rule that every point on the circle satisfies, and no other point does. Because a circle is the set of points a fixed distance r from a centre $(h, k)$, the equation simply says "the distance from $(x, y)$ to the centre equals r."
In its cleanest standard form, the equation of a circle with centre $(h, k)$ and radius r is:
$$(x - h)^2 + (y - k)^2 = r^2.$$
When the centre sits at the origin $(0, 0)$, the equation collapses to its simplest version:
$$x^2 + y^2 = r^2.$$
That is the whole idea. The two numbers $h$ and $k$ slide the circle around the plane, and $r$ sets its size. Read the equation and you read the circle.
Deriving the Equation from the Pythagorean Theorem
The equation is not handed down from nowhere; it comes straight out of the Pythagorean theorem and the distance formula. Take any point $(x, y)$ on a circle centred at $(h, k)$. Drop a horizontal line and a vertical line to form a right triangle with the radius as the hypotenuse:
The horizontal leg has length $x - h$ (the difference in x-coordinates).
The vertical leg has length $y - k$ (the difference in y-coordinates).
The hypotenuse is the radius, $r$.
The Pythagorean theorem says hypotenuse squared equals the sum of the squares of the legs:
$$(x - h)^2 + (y - k)^2 = r^2.$$
That is the standard form, derived in one line. The distance formula you may have met earlier is the same statement: the distance from $(x, y)$ to $(h, k)$ is $\sqrt{(x-h)^2 + (y-k)^2}$, and setting that distance equal to $r$ and squaring both sides gives the equation again.
The General Form of the Equation of a Circle
Expand the standard form and you get a second, equivalent way of writing the same circle, called the general form:
$$x^2 + y^2 + 2gx + 2fy + c = 0.$$
Here the constants relate to the centre and radius by:
$$\text{centre} = (-g, -f), \qquad r = \sqrt{g^2 + f^2 - c}.$$
The general form is what you usually meet first in a problem, because expanding hides the centre and radius. The standard form puts them on display; the general form buries them. So the real skill is converting from one to the other, which is what completing the square does.
How to Find the Centre and Radius — Completing the Square
To pull the centre and radius out of a general-form equation, complete the square on the x-terms and the y-terms separately. The move is always the same: take half the coefficient of x, square it, and add it (and do the same for y), balancing the equation as you go. Group the x-terms and y-terms, complete each square, and the equation reorganises into standard form, where the centre and radius read off directly. Example 4 below walks through the full procedure on real numbers.
Examples of the Equation of a Circle
With both forms, the derivation, and the conversion in hand, here is the equation doing real work, from reading a centre off the standard form up to building an equation from three given points. The problems build from clean reading to a completing-the-square conversion.
Example 1 - Find the centre and radius of the circle $(x - 4)^2 + (y + 7)^2 = 9$
Match against $(x - h)^2 + (y - k)^2 = r^2$. Here $h = 4$, and since $(y + 7) = (y - (-7))$, we have $k = -7$. The right side is $r^2 = 9$, so $r = 3$.
Final answer: centre $(4, -7)$, radius 3.
Example 2 - Write the equation of a circle with centre $(-2, 5)$ and radius 6
A first instinct is to drop the centre coordinates straight into the equation with their own signs and write $(x + (-2))^2 + (y + 5)^2 = 6$, making two sign and squaring errors at once. Check the form: the standard equation subtracts the centre coordinate, $(x - h)$, and the right side is $r^2$, not $r$. With $h = -2$, the term is $(x - (-2)) = (x + 2)$; with $k = 5$, it is $(y - 5)$; and the right side is $6^2 = 36$.
The correct equation is:
$$(x + 2)^2 + (y - 5)^2 = 36.$$
Final answer: $(x + 2)^2 + (y - 5)^2 = 36$.
Example 3 - Write the equation of a circle centred at the origin with radius 10
With centre $(0, 0)$ the equation is just $x^2 + y^2 = r^2$:
$$x^2 + y^2 = 100.$$
Final answer: $x^2 + y^2 = 100$.
Example 4 - Find the centre and radius of the circle $x^2 + y^2 - 6x + 4y - 12 = 0$
Group and complete the square on x and y separately. For x: half of $-6$ is $-3$, squared is 9. For y: half of $4$ is $2$, squared is 4. Add 9 and 4 to both sides:
$$(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4.$$
$$(x - 3)^2 + (y + 2)^2 = 25.$$
Reading off the standard form: centre $(3, -2)$, and $r = \sqrt{25} = 5$.
Final answer: centre $(3, -2)$, radius 5.
Example 5 - Find the centre and radius of $x^2 + y^2 + 8x - 10y + 5 = 0$ using the general-form shortcut
Match against $x^2 + y^2 + 2gx + 2fy + c = 0$: here $2g = 8$ so $g = 4$; $2f = -10$ so $f = -5$; and $c = 5$. Then:
$$\text{centre} = (-g, -f) = (-4, 5), \qquad r = \sqrt{g^2 + f^2 - c} = \sqrt{16 + 25 - 5} = \sqrt{36} = 6.$$
Final answer: centre $(-4, 5)$, radius 6.
Example 6 - A circle has a diameter with endpoints $A(1, 2)$ and $B(7, 10)$. Find its equation
The centre is the midpoint of the diameter, and the radius is half the diameter's length. Midpoint:
$$\left(\frac{1 + 7}{2}, \frac{2 + 10}{2}\right) = (4, 6).$$
Diameter length $= \sqrt{(7-1)^2 + (10-2)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$, so $r = 5$. The equation is:
$$(x - 4)^2 + (y - 6)^2 = 25.$$
Final answer: $(x - 4)^2 + (y - 6)^2 = 25$.
Where the Equation of a Circle Shows Up
The equation matters because it turns a circle into something a computer or an engineer can calculate with, not just draw.
GPS and trilateration. Each satellite tells your phone it is some distance away, which is a circle (a sphere in 3D) centred on that satellite. Your location is where these circles intersect, found by solving their equations together. Without the circle equation there is no GPS fix.
Computer graphics and games. Every rendered wheel, planet, or collision boundary is stored as a centre and radius; the equation decides whether a pixel or a moving object is inside, on, or outside the circle.
Engineering and CAD. Designing a circular gear, a pipe cross-section, or a roundabout starts from the circle's equation in a coordinate system, so the part can be machined to exact tolerances.
Conic sections and orbits. The circle is the simplest conic section, and its equation is the gateway to the ellipse, parabola, and hyperbola that describe planetary orbits and satellite dishes.
For a Grade 11 student, the equation of a circle is the first real bridge between algebra and geometry, the moment a shape becomes an equation you can manipulate, which is exactly the leap that coordinate geometry and later calculus depend on.
Where Students Trip Up on the Equation of a Circle
Mistake 1: Getting the sign wrong on the centre
Where it slips in: A circle is centred at $(-3, 4)$ and the student writes $(x - 3)^2$ or $(y + 4)^2$, copying the centre's signs straight across.
Don't do this: Treat the centre coordinate's sign as the sign inside the bracket.
The correct way: The form subtracts the centre, $(x - h)$. For $h = -3$, that becomes $(x - (-3)) = (x + 3)$. The sign in the bracket is the opposite of the centre coordinate's sign. The rusher who copies signs is the one this catches.
Mistake 2: Writing r instead of r² on the right
Where it slips in: A student knows the radius is 5 and writes the right-hand side as 5 instead of 25.
Don't do this: Put $r$ on the right of the equation.
The correct way: The standard form ends in $r^2$, so a radius of 5 gives a right side of $25$. When reading an equation in reverse, take the square root: a right side of 49 means $r = 7$, not 49.
Mistake 3: Forgetting to balance both sides when completing the square
Where it slips in: A student adds the completing-the-square constants to the left side only.
Don't do this: Add 9 and 4 to the left without adding them to the right.
The correct way: Whatever you add to complete the square on the left must also be added to the right, to keep the equation true. Skip that and the radius comes out wrong. The memorizer who recalls "complete the square" without the balancing step lands on the wrong $r^2$.
Key Takeaways
The equation of a circle in standard form is $(x - h)^2 + (y - k)^2 = r^2$, with centre $(h, k)$ and radius $r$.
It is derived directly from the Pythagorean theorem: the radius is the hypotenuse of a right triangle with legs $(x - h)$ and $(y - k)$.
The general form $x^2 + y^2 + 2gx + 2fy + c = 0$ hides the centre $(-g, -f)$ and radius $\sqrt{g^2 + f^2 - c}$.
Complete the square to convert general form to standard form, balancing both sides as you go.
The most common mistakes are flipping the centre's sign and writing $r$ instead of $r^2$.
Practice These Problems to Solidify Your Understanding
Find the centre and radius of $(x - 5)^2 + (y + 3)^2 = 49$.
Write the equation of a circle with centre $(0, -4)$ and radius 8.
Find the centre and radius of $x^2 + y^2 - 4x + 6y - 3 = 0$.
Answer to Question 1: centre $(5, -3)$, radius 7. Answer to Question 2: $x^2 + (y + 4)^2 = 64$. Answer to Question 3: centre $(2, -3)$, radius 4. If Question 1 gave you a radius of 49, you forgot to take the square root of the right-hand side (see Mistake 2).
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