How do you graph #(x+5)^2+(y-2)^2=9#?

Question

Violet Aldrich · Accepted Answer

This is a circle with centre #(-5, 2)# and radius #3#

Given:

#(x+5)^2+(y-2)^2=9#

This can be rewritten slightly as:

#(x-(-5))^2+(y-2)^2 = 3^2#

which is in the standard form:

#(x-h)^2+(y-k)^2=r^2#

which is the equation of a circle with centre #(h, k) = (-5, 2)# and radius #r=3#.

graph{((x+5)^2+(y-2)^2-9)((x+5)^2+(y-2)^2-0.02) = 0 [-15.75, 4.25, -3.12, 6.88]}

James Atkins · Answer

undefined To graph the equation $ (x+5)^2 + (y-2)^2 = 9 $, which represents a circle, follow these steps:

1. Identify the center of the circle from the equation $ (x-h)^2 + (y-k)^2 = r^2 $, where (h, k) is the center of the circle and r is the radius. In this case, the center is (-5, 2).

2. Determine the radius of the circle. In the equation $ (x+5)^2 + (y-2)^2 = 9 $, the radius is 3 because $ r^2 = 9 $ implies $ r = 3 $.

3. Plot the center of the circle at (-5, 2) on the coordinate plane.

4. Using the radius of 3 units, mark points on the circle. You can do this by going 3 units to the right, left, up, and down from the center to find the points on the circle.

5. Connect these points smoothly to draw the circle.

6. Label the center and any other relevant points, such as the points where the circle intersects the axes if applicable.

By following these steps, you'll accurately graph the circle represented by the equation $ (x+5)^2 + (y-2)^2 = 9 $ on the coordinate plane.