How do you write the equation of a circle?

To write the equation of a circle, you only need two pieces of information: the coordinates of its center and the length of its radius. Once you have those, you plug them into the standard circle equation: (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the center and rr is the radius.

Think of a circle like a dog tied to a stake in the yard. The stake is the center (h,k)(h, k), and the tightly pulled leash is the radius rr. The equation simply describes the path the dog makes when it runs all the way around the stake. Every point (x,y)(x, y) on that path is exactly the same distance from the stake.

The Standard Form Equation

The standard form of a circle's equation is (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2. This formula actually comes directly from the Pythagorean theorem (a2+b2=c2a^2 + b^2 = c^2) and the distance formula. If you draw a right triangle connecting the center (h,k)(h, k) to any point on the circle (x,y)(x, y), the horizontal distance is (xh)(x - h), the vertical distance is (yk)(y - k), and the hypotenuse is the radius rr. Because every point on the circle is distance rr from the center, this relationship always holds true.

What if you aren't given the radius?

Sometimes a problem gives you the center of the circle, but instead of the radius, it gives you another point that lies on the circle. To write the equation, you first need to calculate the radius. You can do this by using the distance formula between the center (h,k)(h, k) and the point on the circle (x1,y1)(x_1, y_1). Once you find that distance, you have your rr, and you can square it to finish writing your standard form equation.

Where students slip up

The most common mistake when writing the equation of a circle is messing up the signs for the center coordinates. Because the formula uses minus signs, (xh)(x - h) and (yk)(y - k), the signs in the equation will look opposite to the signs of the center's coordinates. For example, if your center is at (2,5)(-2, 5), the equation will have (x+2)2(x + 2)^2 and (y5)2(y - 5)^2. Another frequent error is forgetting to square the radius at the end of the equation. If your radius is 4, the right side of your equation must be 16, not 4.

Worked through

Write the equation of a circle with a center at (3,4)(-3, 4) that passes through the point (0,8)(0, 8).

Step 1: Identify the center (h,k)(h, k). Here, h=3h = -3 and k=4k = 4.

Step 2: Find the radius rr. The radius is the distance between the center (3,4)(-3, 4) and the point on the circle (0,8)(0, 8). Using the distance formula: r=(0(3))2+(84)2=32+42=9+16=25=5r = \sqrt{(0 - (-3))^2 + (8 - 4)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.

Step 3: Plug hh, kk, and rr into the standard equation (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2. This gives us (x(3))2+(y4)2=52(x - (-3))^2 + (y - 4)^2 = 5^2.

Step 4: Simplify. The final equation is (x+3)2+(y4)2=25(x + 3)^2 + (y - 4)^2 = 25.

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Where this comes from: OpenStax Algebra and Trigonometry, Section 2.2: Rectangular Coordinate System and Graphs · Khan Academy, Conic Sections Unit: Standard equation of a circle

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