Lines and circles: the equation of a circle

A goat is tied to a post with a rope of 5 metres. It can graze anywhere the rope reaches. The edge of that patch of grass is a circle.

Every point on that edge is exactly 5 metres from the post. That is also the maths definition. A circle is all the points at the same distance from one fixed point. The fixed point is the centre M. The distance is the radius r.

In this article you turn that definition into an equation. Then you read the centre and radius off an equation. At the end you make a line cut through a circle.

From Pythagoras to the equation

Take a circle with centre M(1, 1) and radius 5. Pick a point P(x, y) on the circle. Walk from M horizontally until you are straight below P. Then walk straight up to P. This gives a right triangle.

• The horizontal side is x − 1.
• The vertical side is y − 1.
• The hypotenuse is the radius: 5.

Pythagoras then gives:

(x − 1)² + (y − 1)² = 5² = 25

Take P(4, 5). Then x − 1 = 3 and y − 1 = 4. You get 3² + 4² = 9 + 16 = 25. So P is on the circle.

Drag P around the circle

• Circle (x − 1)² + (y − 1)² = 25
• Radius MP = 5
• Shorter sides

Point P

(4, 5)

Horizontal side

x − 1 = 3

Vertical side

y − 1 = 4

Pythagoras

3² + 4² = 9 + 16 = 25

Show the numbers

Point P	x − 1	y − 1	(x − 1)² + (y − 1)²
(−3, −2)	−4	−3	16 + 9 = 25
(−2, −3)	−3	−4	9 + 16 = 25
(1, −4)	0	−5	0 + 25 = 25
(4, −3)	3	−4	9 + 16 = 25
(5, −2)	4	−3	16 + 9 = 25
(6, 1)	5	0	25 + 0 = 25
(5, 4)	4	3	16 + 9 = 25
(4, 5)	3	4	9 + 16 = 25
(1, 6)	0	5	0 + 25 = 25
(−2, 5)	−3	4	9 + 16 = 25
(−3, 4)	−4	3	16 + 9 = 25
(−4, 1)	−5	0	25 + 0 = 25

If P is to the left of M, then x − 1 is negative. Its square is still positive. So the equation works for every point on the circle.

This holds for every circle. With centre M(a, b) and radius r, the equation of the circle is:

(x − a)² + (y − b)² = r²

Reading off the centre and radius

You can also read the equation the other way round. Take this circle:

(x − 2)² + (y + 1)² = 13

Next to x is − 2, so a = 2. Next to y is + 1. Write that as y − (−1) and you see b = −1. The centre is M(2, −1).

On the right is r² = 13. So the radius is r = √13 ≈ 3.61.

Drag the centre, slide the radius

(x − 2)² + (y + 1)² = 16

• Circle
• Radius

r = 4

Centre

M(2, −1)

Radius

r = 4, r² = 16

Without brackets

x² + y² − 4x + 2y − 11 = 0

Show the numbers

Point	(x − 2)² + (y + 1)²
(6, −1)	4² + 0² = 16
(2, 3)	0² + 4² = 16
(−2, −1)	(−4)² + 0² = 16
(2, −5)	0² + (−4)² = 16

On, inside or outside the circle

The left-hand side of the equation is the square of the distance to M. Substitute a point and compare the result with r².

• If the result equals r², the point is on the circle.
• If the result is less than r², the point is inside the circle.
• If the result is greater than r², the point is outside the circle.

Take (x − 2)² + (y + 1)² = 13 again.

Q(4, 2): (4 − 2)² + (2 + 1)² = 4 + 9 = 13, so on the circle
Q(3, 0): (3 − 2)² + (0 + 1)² = 1 + 1 = 2, so inside the circle

Drag Q around the grid below. Eight grid points lie on this circle. A grid point has whole numbers as coordinates. Find all eight. A hint: 13 = 4 + 9.

On, inside or outside

• Circle (x − 2)² + (y + 1)² = 13
• Inside the circle
• Point Q

Point Q

(4, 2)

Substitute

(4 − 2)² + (2 + 1)² = 4 + 9 = 13

Compare

13 = 13: on the circle

Grid points found

1 of 8

Show the numbers

Point Q	(x − 2)² + (y + 1)²	Position
(4, 2)	13	on the circle
(3, 0)	2	inside the circle
(6, 1)	20	outside the circle
(0, −4)	13	on the circle
(−2, −1)	16	outside the circle

Completing the square

Sometimes the equation comes without brackets:

x² + y² − 4x + 2y − 8 = 0

You cannot read M and r off this directly. First you put the equation back in the form with brackets. This is called completing the square.

You can see the idea in a square with side x. Its area is x². Take away a strip 2 wide on the right. Take away a strip 2 wide along the top as well. Together that is 4x. You have now taken away the 2 by 2 corner twice. Put it back once. A square with side x − 2 is left.

Completing the square with areas

• Square (x − 2)²
• Strips 2 wide, 4x together
• 2 by 2 corner, taken away twice

Side: x = 6

x² − 4x + 4

36 − 24 + 4 = 16

(x − 2)²

4² = 16

So

x² − 4x + 4 = (x − 2)²

Show the numbers

x	x²	4x	x² − 4x + 4	(x − 2)²
4	16	16	4	4
5	25	20	9	9
6	36	24	16	16
7	49	28	25	25
8	64	32	36	36
9	81	36	49	49
10	100	40	64	64

x² − 4x + 4 = (x − 2)²

So x² − 4x = (x − 2)² − 4. The rule: take half the number next to x. That goes inside the brackets. Then subtract its square. It works the same way for y. Half of 2 is 1.

x² − 4x = (x − 2)² − 4
y² + 2y = (y + 1)² − 1

Put this into the equation:

(x − 2)² − 4 + (y + 1)² − 1 − 8 = 0
(x − 2)² + (y + 1)² − 13 = 0
(x − 2)² + (y + 1)² = 13

The centre is M(2, −1) and the radius is √13. That is the same circle as above. If a negative number ends up on the right, there is no circle.

A line and a circle

A line and a circle share two points, one point or none. Take the circle x² + y² = 25. Its centre is the origin O(0, 0) and its radius is 5. Slide the line y = x + b up and down below.

Slide the line y = x + b

b = 1

x² + y² = 25 and y = x + 1

• Circle x² + y² = 25
• Line y = x + b
• Shared points

Substituted

2x² + 2x − 24 = 0

Discriminant

D = 2² − 4 × 2 × (−24) = 4 + 192 = 196

Number of points

two intersection points

Shared points

(−4, −3) and (3, 4)

Show the numbers

b	D	Number of points
−8	−56	0
−5√2 ≈ −7.07	0	1
−7	4	2
−6	56	2
−5	100	2
−4	136	2
−3	164	2
−2	184	2
−1	196	2
0	200	2
1	196	2
2	184	2
3	164	2
4	136	2
5	100	2
6	56	2
7	4	2
5√2 ≈ 7.07	0	1
8	−56	0

Take the line y = x + 1. Replace y in the equation of the circle with x + 1:

x² + (x + 1)² = 25
x² + x² + 2x + 1 = 25
2x² + 2x − 24 = 0
x² + x − 12 = 0
(x + 4)(x − 3) = 0
x = −4 or x = 3

For x = 3, y = 3 + 1 = 4. For x = −4, y = −4 + 1 = −3. The intersection points are (3, 4) and (−4, −3).

Substituting always gives a quadratic equation. The discriminant D tells you how many solutions it has.

• If D is positive, there are two intersection points.
• If D is zero, there is one point. The line touches the circle and is called a tangent.
• If D is negative, the line misses the circle.

For y = x + b, substituting gives 2x² + 2bx + b² − 25 = 0. Then D = 200 − 4b². D is zero when b² = 50, so at b = 5√2 ≈ 7.07 and at b = −5√2.

The tangent is perpendicular to the radius

A tangent touches the circle at one point P. The radius to P is always perpendicular to the tangent. Drag P around the circle below.

Drag the point of contact P

• Circle x² + y² = 25
• Radius OP
• Tangent

Point of contact P

(3, 4)

Slope of OP

4/3

Slope of the tangent

−3/4

Product

4/3 × (−3/4) = −1

Tangent

3x + 4y = 25

Show the numbers

Point of contact P	Slope of OP	Slope of the tangent	Tangent
(−4, −3)	3/4	−4/3	−4x − 3y = 25
(−3, −4)	4/3	−3/4	−3x − 4y = 25
(0, −5)	undefined, the line is vertical	0	y = −5
(3, −4)	−4/3	3/4	3x − 4y = 25
(4, −3)	−3/4	4/3	4x − 3y = 25
(5, 0)	0	undefined, the line is vertical	x = 5
(4, 3)	3/4	−4/3	4x + 3y = 25
(3, 4)	4/3	−3/4	3x + 4y = 25
(0, 5)	undefined, the line is vertical	0	y = 5
(−3, 4)	−4/3	3/4	−3x + 4y = 25
(−4, 3)	−3/4	4/3	−4x + 3y = 25
(−5, 0)	0	undefined, the line is vertical	x = −5

Two lines are perpendicular when the product of their slopes is −1. With that you find the tangent to x² + y² = 25 at P(3, 4).

slope of OP = 43
slope of the tangent = −34, because 43 × −34 = −1

The tangent is y = −34x + c. Substitute P(3, 4) to find c:

4 = −34 × 3 + c = −94 + c
c = 4 + 94 = 254

Multiply y = −34x + 254 by 4. You get 4y = −3x + 25, which is 3x + 4y = 25.

Check it with the discriminant. Substituting into the circle gives x² − 6x + 9 = 0. Then D = 36 − 36 = 0. There is exactly one shared point: (3, 4).

At P(5, 0) the radius is horizontal. The tangent is then vertical: x = 5. A vertical line has no slope.