How to center a label on a specific position?

Graph the circle with center at (-1, 2) and diameter 12. Label the center and at least four points on the circ?

• 1. Graph the circle with center at (-1, 2) and diameter 12. Label the center and at least four points on the circle. Write the equation of the circle. 2. Graph the circle with center at (3, -2), which also passes through the point (0, 2). Label the center and at least four points on the circle. Write the equation of the circle. 3.Graph the circle with center at (-1, -2), which also passes through the point (0, 2). Label the center and at least four points on the circle. Write the equation of the circle. 4. Challenge: Graph the equation with a diameter that has endpoints at (-3, 4) and (5, -2). Label the center and at least four points on the circle. Write the equation of the circle. 5. Challenge: Graph the equation with a diameter that has endpoints at (-2, -2) and (4, 2). Label the center and at least four points on the circle. Write the equation of the circle.

• Answer:

  5. Graph the equation with a diameter with endpoints (-2, -2) and (4, 2). First off, since (-2, -2) and (4, 2) comprise the diameter of the circle, calculate the distance using the distance formula. d = sqrt ( (x2 - x1)^2 + (y2 - y1)^2 ). Let (x1, y1) = (-2, -2) and (x2, y2) = (4, 2). d = sqrt( (4 - (-2))^2 + (2 - (-2))^2 ) d = sqrt( (4 + 2)^2 + (2 + 2)^2 ) d = sqrt( 6^2 + 4^2 ) d = sqrt( 36 + 16) d = sqrt(52) d = sqrt(4 * 13) d = 2sqrt(13) So the distance (and therefore diameter) is equal to d = sqrt(52). Next, we need to find the center of the circle. The center is simply the average of the x-coordinates, and the average of the y-coordinates. Let's call (h, k) the coordinates of the circle. h = (x1 + x2)/2 = (-2 + 4)/2 = 2/2 = 1 k = (y1 + y2)/2 = (-2 + 2)/2 = 0/2 = 0 So the center is located at (1, 0). The formula of a circle is (x - h)^2 + (y - k)^2 = r^2 So we want the radius r. Since the diameter is 2sqrt(13), and the radius is half the diameter, it follows that r = (1/2) of diamater r = (1/2) (2sqrt(13)) r = sqrt(13) Making the formula (x - h)^2 + (y - k)^2 = r^2 Given that r = sqrt(13), (h, k) = (1, 0), we get (x - 1)^2 + (y - 0)^2 = [sqrt(13)]^2 (x - 1)^2 + y^2 = 13