Unit 5, lesson 54

Circle Equations, Centers and Radii

How the equation of a circle comes from the distance formula, how to read its center and radius, and how to recover them by completing the square or from the endpoints of a diameter.

  • Theory part 107
  • Practice part 108
  • Circles

Theory, part 107

Practice, part 108

A sprinkler fixed at one spot waters every point within five meters of it, and the edge of the wet patch is a circle. On a coordinate grid that edge is the set of points exactly five units from the sprinkler, and writing “exactly five units away” with the distance formula gives the circle’s equation. Once you see where the equation comes from, reading off a center and radius, or building an equation from scratch, takes only a few lines.

See also Completing the Square and Circles, Semicircles, and Choosing a Branch in Foundations for Calculus and Circles, Distance, and Midpoints in Precalculus.

Key terms

Distance formula

Definition The distance between the points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) in the coordinate plane is

d=(x2−x1)2+(y2−y1)2.d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

It is the Pythagorean theorem applied to the right triangle whose legs are the horizontal change x2−x1x_2 - x_1 and the vertical change y2−y1y_2 - y_1. Neither this formula nor the circle equation is on the SAT reference sheet, so both are worth knowing by heart.

In plain words Go across, go up, and the straight path between the two points is the hypotenuse.

Example From (1,2)(1, 2) to (7,10)(7, 10) the horizontal change is 66 and the vertical change is 88, so

d=62+82=100=10.d = \sqrt{6^2 + 8^2} = \sqrt{100} = 10.

Center and radius

Definition A circle is the set of all points in a plane at a fixed positive distance rr from a fixed point. The fixed point is the center and the distance rr is the radius. A segment from the center to a point of the circle is also called a radius.

In plain words The center is the pin and the radius is the length of string. Every point of the circle is exactly that far from the pin, no closer and no farther.

Example The point (5,1)(5, 1) is on the circle with center (2,−3)(2, -3) and radius 55, because its distance from the center is

(5−2)2+(1+3)2=9+16=5.\begin{aligned} \sqrt{(5 - 2)^2 + (1 + 3)^2} &= \sqrt{9 + 16} \\[0.4em] &= 5. \end{aligned}

The point (2,1)(2, 1) is not, since its distance from the center is only 44.

Circle equation

Definition A point (x,y)(x, y) lies on the circle with center (h,k)(h, k) and radius r>0r > 0 exactly when its distance from (h,k)(h, k) is rr. Squaring both sides of that distance condition gives the standard form

(x−h)2+(y−k)2=r2.(x - h)^2 + (y - k)^2 = r^2.

The constant on the right is the radius squared, not the radius. Signs inside the parentheses are subtractions, so (x+4)2(x + 4)^2 means h=−4h = -4. To see a true circle on a graph, the axes must use the same scale, since unequal scales stretch it into an oval.

In plain words Read the center by flipping the signs you see next to xx and yy. Read the radius by taking the square root of the number on the right.

Example For (x+4)2+(y−1)2=49(x + 4)^2 + (y - 1)^2 = 49, write x+4x + 4 as x−(−4)x - (-4). The center is (−4,1)(-4, 1) and the radius is 49=7\sqrt{49} = 7. Two classic wrong answers are the center (4,−1)(4, -1) and the radius 4949.

Building and moving a circle

Definition To write the equation of a circle from its center (h,k)(h, k) and one point (x1,y1)(x_1, y_1) on it, compute

r2=(x1−h)2+(y1−k)2r^2 = (x_1 - h)^2 + (y_1 - k)^2

and substitute into standard form. Moving a circle aa units right and bb units up moves its center to (h+a,k+b)(h + a, k + b) and leaves the radius unchanged, so the right side r2r^2 stays the same.

In plain words The center and one point give you the string length. Sliding the circle moves the pin but does not change the string.

Example The circle with center (2,−3)(2, -3) that passes through (5,1)(5, 1) has r2=9+16=25r^2 = 9 + 16 = 25, so its equation is (x−2)2+(y+3)2=25(x - 2)^2 + (y + 3)^2 = 25. Check the point in the equation. The left side is 32+42=253^2 + 4^2 = 25, as it should be. Moving this circle 33 units right and 22 units up puts the center at (5,−1)(5, -1) and gives (x−5)2+(y+1)2=25(x - 5)^2 + (y + 1)^2 = 25.

Completing the square

Definition Expanding standard form gives an equation like x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0. To return to standard form, complete the square separately in xx and in yy, using

x2+Dx=(x+D2)2−D24,x^2 + Dx = \left(x + \frac{D}{2}\right)^2 - \frac{D^2}{4},

and the same identity for yy. Whatever you add to one side you must add to the other. The result has the form (x−h)2+(y−k)2=c(x - h)^2 + (y - k)^2 = c. If c>0c > 0, the graph is a circle with radius c\sqrt{c}. If c=0c = 0, the only solution is the single point (h,k)(h, k). If c<0c < 0, no real point satisfies the equation, because a sum of squares of real numbers is never negative.

In plain words Keep the xx terms together and the yy terms together, turn each pair into a perfect square, and move every leftover constant to the right side.

Example Rewrite x2+y2−4x+6y−12=0x^2 + y^2 - 4x + 6y - 12 = 0. Move the constant to the right and group the xx terms and the yy terms. Adding 44 makes x2−4x+4=(x−2)2x^2 - 4x + 4 = (x - 2)^2, and adding 99 makes y2+6y+9=(y+3)2y^2 + 6y + 9 = (y + 3)^2. Both numbers go on the right side too.

(x−2)2+(y+3)2=12+4+9=25.\begin{aligned} (x - 2)^2 + (y + 3)^2 &= 12 + 4 + 9 \\[0.4em] &= 25. \end{aligned}

The center is (2,−3)(2, -3) and the radius is 55. With +13+13 in place of −12-12 the right side would be 00 and the graph a single point, and with +20+20 it would be −7-7 and there would be no graph at all.

Normalized circle equation

Definition An equation ax2+ay2+Dx+Ey+F=0ax^2 + ay^2 + Dx + Ey + F = 0 in which x2x^2 and y2y^2 have the same nonzero coefficient aa can be normalized by dividing every term by aa. This gives an equivalent equation with leading coefficients 11, which is ready for completing the square. If the coefficients of x2x^2 and y2y^2 are different, the graph is not a circle.

In plain words Make the squared terms plain x2x^2 and y2y^2 before you complete any squares, or the numbers you add will be wrong.

Example Start from 2x2+2y2+8x−12y+8=02x^2 + 2y^2 + 8x - 12y + 8 = 0. Dividing by 22 gives x2+y2+4x−6y+4=0x^2 + y^2 + 4x - 6y + 4 = 0. Now add 44 and 99 to both sides after moving the 44.

(x+2)2+(y−3)2=−4+4+9=9.\begin{aligned} (x + 2)^2 + (y - 3)^2 &= -4 + 4 + 9 \\[0.4em] &= 9. \end{aligned}

The center is (−2,3)(-2, 3) and the radius is 33.

Diameter endpoints

Definition A diameter is a chord through the center, and its length is 2r2r. If (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are the endpoints of a diameter, the center is their midpoint

(h,k)=(x1+x22,y1+y22),(h, k) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right),

and the radius is half the distance between them.

In plain words The center sits exactly halfway along any diameter, and the radius is half its length.

Example A diameter runs from (−1,4)(-1, 4) to (7,−2)(7, -2). The center is (62,22)=(3,1)\left(\frac{6}{2}, \frac{2}{2}\right) = (3, 1). The diameter has length 82+62=10\sqrt{8^2 + 6^2} = 10, so the radius is 55, and the circle is (x−3)2+(y−1)2=25(x - 3)^2 + (y - 1)^2 = 25. Both endpoints check, since (−4)2+32=25(-4)^2 + 3^2 = 25 and 42+(−3)2=254^2 + (-3)^2 = 25.

Check yourself

  1. The equation x2+y2+10x−4y=7x^2 + y^2 + 10x - 4y = 7 defines a circle in the xyxy-plane. What is the radius of the circle? (A) 7\sqrt{7}, (B) 29\sqrt{29}, (C) 66, (D) 3636
Show the answer

The answer is (C). Adding 2525 and 44 to both sides gives

(x+5)2+(y−2)2=36,(x + 5)^2 + (y - 2)^2 = 36,

so the radius is 36=6\sqrt{36} = 6. Choice (D) is the radius squared, choice (A) forgets to add the two square terms, and choice (B) adds them but drops the 77.

  1. A circle in the xyxy-plane has a diameter with endpoints (2,3)(2, 3) and (8,11)(8, 11). An equation of the circle is (x−h)2+(y−k)2=c(x - h)^2 + (y - k)^2 = c, where hh, kk and cc are constants. What is the value of cc?
Show the answer

The answer is 2525. The center is the midpoint (5,7)(5, 7). The diameter has length 62+82=10\sqrt{6^2 + 8^2} = 10, so the radius is 55 and c=r2=25c = r^2 = 25. Entering 100100 would use the whole diameter as the radius.

  1. The circle (x−3)2+(y+2)2=16(x - 3)^2 + (y + 2)^2 = 16 is moved 55 units to the left and 11 unit up. Which equation describes the moved circle? (A) (x−8)2+(y+3)2=16(x - 8)^2 + (y + 3)^2 = 16, (B) (x+2)2+(y+1)2=16(x + 2)^2 + (y + 1)^2 = 16, (C) (x+2)2+(y+3)2=16(x + 2)^2 + (y + 3)^2 = 16, (D) (x+2)2+(y+1)2=4(x + 2)^2 + (y + 1)^2 = 4
Show the answer

The answer is (B). The center moves from (3,−2)(3, -2) to (3−5,−2+1)=(−2,−1)(3 - 5, -2 + 1) = (-2, -1) and the radius stays 44, so the right side stays 1616. Choice (A) moves the center the wrong way in both directions, choice (C) moves it down instead of up, and choice (D) writes the radius where the radius squared belongs.

How this lesson connects