There are four types of conic sections that students will learn in an advanced algebra course. One of which is the circle. Suppose you have a point and place another point 2 inches from it. Now return to the original point and place another point 2 inches from it. As you continue to repeat this process, the points that are equidistant from the original point will get closer and closer together. Eventually this will start to resemble a circle.
Recall that the Pythagorean Theorem states that in a right triangle with sides of length a, b and hypotenuse c , a2 + b2 = c2 . Solving for c, we get c2 = √(a2 + b2 ). We can continue with this concept to obtain the equation of a circle. If we let the coordinate (x , y ) be a coordinate on a circle with the center (h , k ) , the distance between these two points is called the radius r of the circle.
The formula for r is r = √[(x -h )2 + ( y – k )2 ]. If we square both sides of the equation to remove the radical we get the equation of a circle in standard form, which is r2 = (x–h)2 + (y-k)2 .
Examples: Find the center and the radius of each circle and graph it. 1. (x – 2)2 + ( y – 3)2 = 16. Notice that the equation is in standard form, so the center of the circle is (2, 3). We know that r2 = 16, therefore the radius of the circle is 4. To graph the circle put a point at (2, 3) on a rectangular coordinate system. From that point, move 4 units to the left, 4 units to the right, 4 units up, 4 units down and place a point at each. Then draw a circle through the outer points.
Once again the equation is in standard form, so the center of the circle is (-1, 0). We know that r2 = 9, therefore the radius of the circle is 3. Place a point at (-1, 0) on the graph and move 3 units left, right, up, down, and place a point at each. Then draw a circle through the outer points.
3. Write the equation of the circle in standard form with radius 5 and center (4, -1).
We are given all the information we need in the problem, so we substitute 4 for h , -1 for k and 5 for r to get (x – 4)2 + ( y – (-1))2 = 52 .
(x – 4)2 + (y + 1)2 = 52 .
(x – 4)2 + (y + 1)2 = 25.
Oftentimes the equation of a circle is not in standard form. In the previous example, the equation of the circle in standard form was (x – 4)2 + (y + 1)2 = 25. If we square ( x – 4) and ( y + 1) we get the equation in a different, unfactored form called the general form of the equation of the circle. The general form is x2 + y2 + Ax + By + C = 0, where A , B and C are real numbers. We can take the equation in general form and change it to standard form so it is easier to graph. This is done by completing the square.
Example: Write the equation x2 + y2 – 6 x + 4 y – 23 = 0 in standard form and graph it.
First we rewrite the equation grouping the x terms and y terms and adding 23 to both sides. This gives us x2– 6x + y2 + 4y = 23. We now complete the square twice, first with x2 – 6x, then with y2 + 4y. This gives us x2– 6x + 9 + y2 + 4y + 4 = 23 + 9 + 4. Now factor x2– 6x + 9 and y2 + 4y + 4 separately to get (x – 3)2 + (y + 2)2 = 36. Notice that the equation is now in standard form for a circle. We know the radius is 6 and the center is (3, -2).
Example: Suppose you wish to draw three circular archery targets. The first target is given by the equation x 2 + y 2 = 25. The next target is tangent to the first and is centered at (9, 0). The third is tangent to the second one and is centered at (16, 0). What are the equations of the other circles?
First we have to figure out the center and radius of the first circle. The equation of the circle is in standard form s o the center is (0, 0). We know that r2 = 25, therefore r = 5 and an endpoint is at (5, 0).
Since the second circle is tangent to the first circle at (5, 0) and the center is at (9, 0), that means the radius of the second circle is 4. So we know that h = 9, k = 0 and r = 4. Therefore the equation of the second circle in standard form is (x – 9)2 + y2 = 16.
Since the third circle is tangent to the second circle at the end of the second circle at (13, 0), which is obtained by adding the radius of 4 to the x coordinate of the center. With the center of the third circle at (16, 0), the radius is 3. Therefore, the equation of the third circle in standard form is (x – 16)2 + y2 = 9.
This guide should clear any confusion about equations and graphing of circles.
