MPM2D Equation of a Circle: Notes and Practice

A cell tower stands in the middle of a town. Put the tower at the origin $(0, 0)$ of a map whose units are kilometres. Its signal reaches $5$ km in every direction, so the edge of its coverage is a circle of radius $5$ around the tower. A phone at $(3, 4)$ is exactly on the edge, a phone at $(2, - 3)$ has signal, and a phone at $(4, - 4)$ does not. How can you tell without drawing anything? In Lesson 1 you measured the distance between two points. In this lesson you turn "every point that is $5$ km from the tower" into a single equation, and that equation answers the question for any point you test.

A circle is the set of all points in a plane that are the same distance from a fixed point. The fixed point is the centre, and the fixed distance is the radius, $r$ . A segment through the centre with both endpoints on the circle is a diameter; its length is $2 r$ . In this course every circle you write an equation for has its centre at the origin, $O (0, 0)$ .

A steel lattice cell tower with many antennas rising above green treetops against a clear blue sky. — PhotoA cell tower's signal reaches about the same distance in every direction, so the edge of its coverage is modelled as a circle centred on the tower.David Ratledge, CC BY 4.0, via Wikimedia Commons

Developing the equation

Let $P (x, y)$ be any point on the circle with centre $O (0, 0)$ and radius $r$ . The distance from $O$ to $P$ is $r$ , so the length formula gives $(x - 0)^{2} + (y - 0)^{2} = r$ , which simplifies to $x^{2} + y^{2} = r$ . Squaring both sides removes the square root. The reasoning also runs the other way: if a point's coordinates make $x^{2} + y^{2} = r^{2}$ true, its distance from the origin is $r^{2} = r$ , so it is on the circle. That is what makes this the circle's equation: the points that satisfy it are exactly the points on the circle.

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

Circle with centre (0, 0) and radius r. The right side is the radius squared.

Fig 1The circle x squared plus y squared equals 25, radius 5, centred at the origin. The point P(3, 4) is on it, with legs 3 and 4 and hypotenuse 5. Q(2, −3) is inside and R(4, −4) is outside.

The tower's coverage edge has $r = 5$ , so its equation is $x^{2} + y^{2} = 25$ . Test $P (3, 4)$ : $3^{2} + 4^{2} = 9 + 16 = 25$ , so $P$ is on the circle. The legs $3$ and $4$ with hypotenuse $5$ form a right triangle, and every point of the circle that is not on an axis has one like it: the horizontal leg comes from $x$ , the vertical leg comes from $y$ , and the hypotenuse is $r$ . Because the equation only uses $x^{2}$ and $y^{2}$ , changing the sign of a coordinate does not change whether a point fits. So $(- 3, 4)$ , $(3, - 4)$ and $(- 3, - 4)$ are on the circle too, and the circle is symmetric across both axes.

VideoAn Ontario Grade 10 lesson on the equation of a circle centred at the origin, finding its radius, and testing whether points are inside, on or outside it.Watch on YouTube

From equation to radius, and back

The number on the right side of $x^{2} + y^{2} = r^{2}$ is the radius squared. To read the radius, take the positive square root of that number. To write the equation from a radius, square the radius. If the circle passes through a known point $(a, b)$ , substitute the point to get $r^{2} = a^{2} + b^{2}$ directly; you may not even need $r$ itself.

Equation	r squared	Radius r	Decimal
$x^{2} + y^{2} = 49$	$49$	$49 = 7$	$7.0$
$x^{2} + y^{2} = 2.25$	$2.25$	$2.25 = 1.5$	$1.5$
$x^{2} + y^{2} = 20$	$20$	$20 = 25$	$4.5$
$x^{2} + y^{2} = 12$	$12$	$12 = 23$	$3.5$

The radius is the positive square root of the constant on the right. Decimals are rounded to one place.

The right side is r squared, not r

x^{2} + y^{2} = 16

has radius

4

, not

16

. Going the other way, a circle of radius

9

x^{2} + y^{2} = 81

, not

x^{2} + y^{2} = 9

. Square a radical radius with care:

(23)^{2} = 2^{2} \times (3)^{2} = 4 \times 3 = 12 .

To sketch $x^{2} + y^{2} = r^{2}$ , first find $r$ . Plot the four intercepts $(r, 0)$ , $(- r, 0)$ , $(0, r)$ and $(0, - r)$ , add any other points you know (for $r = 5$ , the twelve whole-number points include $(\pm 3, \pm 4)$ and $(\pm 4, \pm 3)$ ), then draw a smooth round curve through them, or use a compass set to the radius. Use the same scale on both axes; with different scales, a circle looks like an oval.

Inside, on or outside

For any point $(x, y)$ , the value of $x^{2} + y^{2}$ is the square of its distance from the origin. Compare it with $r^{2}$ . If $x^{2} + y^{2} < r^{2}$ , the point is closer than $r$ , so it is inside the circle. If $x^{2} + y^{2} = r^{2}$ , it is on the circle. If $x^{2} + y^{2} > r^{2}$ , it is farther than $r$ , so it is outside. Comparing squares is reliable because, for distances (which are never negative), a larger distance always has a larger square.

Point	x squared plus y squared	Compare with 25	Location
$(3, 4)$	$9 + 16 = 25$	$25 = 25$	On the circle
$(2, - 3)$	$4 + 9 = 13$	$13 < 25$	Inside
$(4, - 4)$	$16 + 16 = 32$	$32 > 25$	Outside
$(0, - 5)$	$0 + 25 = 25$	$25 = 25$	On the circle

Testing points against the circle with radius 5 centred at the origin

Only for a centre at the origin

The equation

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

describes a circle centred at

(0, 0)

. A circle with any other centre has a different equation, which you will meet in later courses. Also,

r

must be positive for a real circle:

x^{2} + y^{2} = 0

is satisfied only by the single point

(0, 0)

, and

x^{2} + y^{2} = - 9

has no points at all, since a sum of squares cannot be negative.

Equation of a Circle

Key things to remember

Developing the equation

From equation to radius, and back

Inside, on or outside

Worked examples

Where marks get lost

Flashcards 8

Practice