Equation for a Circle
The equation for a circle whose center is at the origin (0,0) is x2 + y2 = r2. In this case x and y represent
the x and y coordinates on the circle and r represents the radius of the circle, which also represents the
distance from the origin (0,0) to the (x, y) coordinates. It’s really just a use of the Pythagorean
Theorem as it relates to the distance formula. See the example below. In each case the length r is
determined by using the equation x2+ y2 = r2 (looks a lot like a2+ b2= c2). For this example we chose a
radius of 5. In this case x represents the distance from the origin to the x coordinate or x – 0. It also
represents the bottom side of a right triangle. The y represents the distance from the origin to the y
coordinate or y – 0. It also represents the right side of the triangle. The r represents the distance from
the origin to point (x, y), the hypotenuse of the right triangle, or the radius of the circle. The last figure
in the series shows how the points represented by the (x, y) coordinates in each of the first 5 figures,
are all on the same circle.
x=3 , y =4
r=5
x=4 , y =3
r=5
x=5 , y =0
r=5
(-4, 3)
x=2 , y =3.464
r=5
x=0 , y =5
r=5
Even if x or y were negative they would still be on the circle as long as the sum of their squares
equaled r2 in this case 25. For example if one was to choose a value of -4 for x and then used the
equation x2 + y2 = r2. So plugging in the know values we get (-4)2 + y2 = 52. Simplified this would be
16 + y2= 25
y2 = 25 -16
y2 = 9
Find the square root of each side. 
y=3
So the point would be (-4, 3). Notice that the point is on the circle. If one was to determine all cases in
which x2 + y2 = 25, all of the points would be on the above circle.
What about circles that don’t have the origin as their centers? That’s on the next page.
Equation for a Circle
In the previous example for a circle whose center was the origin, the distance from the origin to the x
coordinate on the circle was determined by subtracting the x coordinate of the origin, or x – 0.
Likewise the distance from the origin to the y coordinate on the circle was determined by y – 0. The
zeroes represent the coordinates in the origin (in this case the center of the circle). So to determine the
equation for a circle whose center is a point other than the origin, just replace the zeroes with the
coordinates of the center of the circle. For example if one wants to create a circle whose radius is 5 and
whose center is at (2, 3), use the equation (x – 2)2 + (y – 3)2 = 52.
It will be the same size circle as in the previous example,
but just moved 2 units right and 3 units up.
(2, 3)
A standard way to write this equation is
(x – a)2 + (y – b)2 = r2
where (a, b) represents the coordinates for the center of
the circle, and (x, y) represents the coordinates for a point on
the circle.
Given a segment determine the equation for the circle whose diameter is that segment.
To do this one must first determine the midpoint of the segment because that will be the circle’s center.
For example: Given the segment whose endpoints are (-2, 1) and (4, 5) the midpoint is the point whose
x coordinate is the average of the two x coordinates of the endpoints; and whose y coordinate is the
average of the two y coordinates of the endpoints.
x =
y=
(-2 + 4)/ 2 = 1
(1 + 5)/2 = 3
The midpoint would be (1, 3).
Next we must determine the radius of the circle. Since the segment is the diameter, the radius will be
equal to half of the length of the segment. We can use the distance formula to determine the length of
the segment and then divide that by two to determine the radius. Remember that the distance formula is
(4,5)
just a variation of the Pythagorean Theorem.
The difference between the two x coordinates is 4 –(-2) = 6
(-2, 1)
The difference between the y coordinates is 5 – 1 = 4
The segment is the hypotenuse of a triangle whose height
is the difference between the y coordinates, and whose base
is the difference between the x coordinates. Using the Pythagorean
Theorem the length of the hypotenuse would be 62+ 42= c2 or
36 + 16 = c2  52 = c2  c = 7.2 (the square root of 52).
4
6
The actual distance formula is written as d = ( x2 – x1)2 + (y2 – y1)2

62 + 42
 52

d = 7.2
( 4 - (-2))2 + (5 - 1)2
Remember the radius was half the length of the segment so r = 7.22 or
r = 3.6
Now just plug in the midpoint coordinates and the value for r and we have the equation of the circle.
(x – 1)2+ (y – 3)2 =( 3.6)2
Equation for a Circle
Determine the equations for the following.
1. A circle whose center is at the origin and whose radius is equal to 6.
2. A circle whose center is at the point (3, 7) and whose radius is 4.
3. A circle whose center is at point (-4, 5) and whose radius is 8.
4. Given the segment with endpoints of (3, 4) and (4, 7), determine the equation for the circle that
contains that segment as a diameter.
5. Given the segment with endpoints of (-4, 3) and (1, -5), determine the equation for the circle that
contains that segment as a diameter.
6. Is the point (6, 8) inside, outside, or on the circle defined by the equation x2+ y2 = 100? Show your
work.
Hint: determine how far the point is away from the center of the circle. If it is less than the radius
then it is in the circle; if it is greater than the radius then it is outside the circle; if it is the same as the
radius then it is on the circle.
7. Is the point (4, 5) inside, outside, or on the circle defined by the equation (x – 1)2+ (y – 2)2 = 16?
Show your work. Hint: Find the center of the circle from the equation. The x coordinate for the center
is 1.