The Equation of a Circle

Pythagoras from the center, written down.

Every point the same distance from the center

A circle is the set of all the points that are the same distance from one fixed point, its center. That distance is the radius, r.

On a coordinate grid, call the center (a, b) and any point on the circle (x, y). Whichever point of the circle you choose, its distance from (a, b) is r. The equation of the circle is the condition on x and y that says exactly this.

xy(a, b)(x, y)

The arrow is a radius, from the center (a, b) to a point (x, y) on the circle. Every point of the circle is the same distance r from the center.

Across, then up

Go from the center to the point in two moves: across, parallel to the x-axis, and then up, parallel to the y-axis. The distance across is the difference between the x-coordinates, x − a. The distance up is the difference between the y-coordinates, y − b.

These two moves meet at a right angle, so together with the radius they make a right triangle. Its two shorter sides are x − a and y − b, and its longest side, the hypotenuse, is the radius r.

xyx − ay − b(a, b)

Across x − a and up y − b, from the center to the point on the circle: the two shorter sides of a right triangle whose hypotenuse is the radius.

Pythagoras writes the equation

Pythagoras' theorem says that the squares of the two shorter sides add up to the square of the hypotenuse. For this triangle that is (x − a)² + (y − b)² = r². It holds for every point on the circle and for no other point, so it is the equation of the circle with center (a, b) and radius r.

In the drawing the center is (2, 2) and the radius is 5, so the circle is (x − 2)² + (y − 2)² = 25. Test it on the point (5, 6): it is 5 − 2 = 3 across and 6 − 2 = 4 up, and 3² + 4² = 9 + 16 = 25. Test it on (−1, 6) too: (−1 − 2)² + (6 − 2)² = (−3)² + 4² = 9 + 16 = 25. The move across is −3 there, to the left, but squaring removes the sign, so the direction of each move does not matter.

A point that is not on the circle gives a different number. For (4, 3), (4 − 2)² + (3 − 2)² = 4 + 1 = 5, which is less than 25. So (4, 3) is √5 from the center, less than the radius of 5, and it is inside the circle. A point that gives more than r² is outside it.

A circle centered at the origin

When the center is the origin, a = 0 and b = 0, and the brackets become (x − 0)² and (y − 0)², which are just x² and y². So a circle centered at the origin has the equation x² + y² = r².

For example, x² + y² = 9 is the circle centered at the origin with radius 3, since 3² = 9. It passes through (3, 0), (0, 3), (−3, 0) and (0, −3): each of these gives 9 + 0 or 0 + 9.

xy

x² + y² = 9: center the origin, radius 3.

Reading the center and the radius

Given an equation in this form, read it backwards. Take (x + 1)² + (y − 2)² = 9. The general form subtracts the center's coordinates, and x + 1 is x − (−1), so a = −1. The second bracket is y − 2, so b = 2. The center is (−1, 2): each coordinate has the opposite sign to the number in its bracket.

The number on the right is r², not r. Here r² = 9, so the radius is √9 = 3.

Writing an equation works the same way in the other direction. The circle with center (3, −4) and radius 2 is (x − 3)² + (y − (−4))² = 2², which tidies to (x − 3)² + (y + 4)² = 4.

xy(−1, 2)

(x + 1)² + (y − 2)² = 9: center (−1, 2), radius 3.

The usual mistakes

Copying the signs from the brackets. (x + 1)² + (y − 2)² = 9 has its center at (−1, 2), not (1, −2). The center is the point that makes both brackets zero.

Taking the number on the right as the radius. In (x + 1)² + (y − 2)² = 9 the radius is √9 = 3, not 9. And a circle of radius 2 has 4 on the right, not 2.

Doubling the radius. The radius reaches from the center to the circle. The distance all the way across is the diameter, 2r.

Where a straight line meets a circle

A common question asks where a straight line crosses a circle. At a point where the line meets the circle, both equations are true, just as at a crossing of two lines. So substitute the line's expression for y into the circle's equation. That leaves an equation in x alone, a quadratic, and each of its roots is the x-coordinate of one meeting point.

For example, the line y = x + 1 meets the circle x² + y² = 25 where x² + (x + 1)² = 25. Expand the bracket: x² + x² + 2x + 1 = 25, so 2x² + 2x − 24 = 0. Divide by 2: x² + x − 12 = 0, which factorizes as (x + 4)(x − 3) = 0. So x = 3 or x = −4, and the line gives y = 4 and y = −3. The meeting points are (3, 4) and (−4, −3). Check: 3² + 4² = 25 and (−4)² + (−3)² = 25.

Worked example: The Range of a Radio Mast, a Town and a Straight Road

Question A radio mast stands at the origin of a map on which 1 unit is 1 km, with x measured to the east and y to the north. The signal reaches every place within 10 km of the mast. (a) A town is at (5, −8). Does the town receive the signal? (b) A straight road follows the line y = x + 2. Find the coordinates of the points where the road enters and leaves the range of the mast.

  1. 1.A point (x, y) is √x2 + y2 km from the mast. On the edge of the range this distance is 10 km, so the edge is the circle x2 + y2 = 102, which is x2 + y2 = 100.

    −10−50510−10−50510km east of the mast, xkm north, ymast10 kmevery point on the edge is 10 km from the mastx2+ y2= 102, so x2+ y2= 100
    −10−50510−10−50510km east of the mast, xkm north, ymast10 kmevery point on the edge is 10 km from the mastx2+ y2= 102, so x2+ y2= 100
    Every point on the edge of the range is 10 km from the mast at the origin, so the edge is the circle x2 + y2 = 100.

  2. 2.For the town at (5, −8), x2 + y2 = 52 + (−8)2 = 25 + 64 = 89. (a) Yes. Since 89 < 100, the town is less than 10 km from the mast, so it is inside the circle and receives the signal.

    −10−50510−10−50510km east of the mast, xkm north, ymasttown (5, −8)10 kmtown: 52+ (−8)2= 25 + 64 = 8989 < 100: the town is inside the circle
    −10−50510−10−50510km east of the mast, xkm north, ymasttown (5, −8)10 kmtown: 52+ (−8)2= 25 + 64 = 8989 < 100: the town is inside the circle
    (a) For the town, 52 + (−8)2 = 25 + 64 = 89. This is less than 100, so the town is inside the circle and receives the signal.

  3. 3.Where the road meets the edge, both equations hold. Substitute y = x + 2 into the circle: x2 + (x + 2)2 = 100. Expand: 2x2 + 4x + 4 = 100. Subtract 100 from both sides and divide both sides by 2: x2 + 2x − 48 = 0.

    −10−50510−10−50510km east of the mast, xkm north, yy = x + 2masttown (5, −8)on the road y = x + 2: x2+ (x + 2)2= 1002x2+ 4x + 4 = 100x2+ 2x − 48 = 0
    −10−50510−10−50510km east of the mast, xkm north, yy = x + 2masttown (5, −8)on the road y = x + 2: x2+ (x + 2)2= 1002x2+ 4x + 4 = 100x2+ 2x − 48 = 0
    Substitute y = x + 2 into the circle: x2 + (x + 2)2 = 100. Expand and divide both sides by 2: x2 + 2x − 48 = 0.

  4. 4.Factorize: two numbers with a product of −48 and a sum of 2 are 8 and −6, so (x + 8)(x − 6) = 0, and x = −8 or x = 6. Substitute each root into y = x + 2: y = −6 when x = −8, and y = 8 when x = 6.

    −10−50510−10−50510km east of the mast, xkm north, ymasttown(−8, −6)(6, 8)(x + 8)(x − 6) = 0, so x = −8 or x = 6y = −8 + 2 = −6, and y = 6 + 2 = 8
    −10−50510−10−50510km east of the mast, xkm north, ymasttown(−8, −6)(6, 8)(x + 8)(x − 6) = 0, so x = −8 or x = 6y = −8 + 2 = −6, and y = 6 + 2 = 8
    Factorize: (x + 8)(x − 6) = 0, so x = −8 or x = 6. The line gives y = −6 and y = 8.

  5. 5.(b) The road enters the range at (−8, −6) and leaves it at (6, 8). Check in the circle: (−8)2 + (−6)2 = 64 + 36 = 100 and 62 + 82 = 36 + 64 = 100.

    −10−50510−10−50510km east of the mast, xkm north, ymasttown(−8, −6)(6, 8)the road enters at (−8, −6) and leaves at (6, 8)check: 64 + 36 = 100, and 36 + 64 = 100
    −10−50510−10−50510km east of the mast, xkm north, ymasttown(−8, −6)(6, 8)the road enters at (−8, −6) and leaves at (6, 8)check: 64 + 36 = 100, and 36 + 64 = 100
    (b) The road enters the range at (−8, −6) and leaves it at (6, 8). Both points satisfy x2 + y2 = 100.

Answer: (a) Yes: 52 + (−8)2 = 89, which is less than 100; (b) at (−8, −6) and (6, 8)

Common mistakes

  • Comparing 89 with 10 and deciding that the town is out of range. The number 89 is the square of the distance, so it must be compared with 102 = 100. The distance itself is √89 ≈ 9.4 km.
  • Expanding (x + 2)2 as x2 + 4. The middle term is missing: (x + 2)2 = x2 + 4x + 4. Without the 4x the quadratic equation has different roots, and the points found are not on the road.

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