Circle
Example 1: Give the standard equation of the circle satisfying the given conditions.
Center at the origin, radius 5
Center (4, 3), radius ?9
Center (-2, 6), radius 4
Solution:
Equation of the circle: (x-h)^2+(y-k)^2=r^2
The center is (0, 0) therefore the equation of the circle is x^2+y^2=25.
The center is (4, 3) and the radius is?9, so the equation is (x-4)^2+(y-3)^2=9.
The center is (-2, 6) and the radius is 4, so the equation is (x+2)^2+(y-6)^2=16.
Example 2: A circle C has a center (2,5) and passes through the point (0, 0). Find the equation for C.
Solution: (x-h)^2+(y-k)^2=r^2
r= (2)^2+(5)^2=29
Therefore the equation of c is: (x-2)^2+(y-5)^2=29.
Example 3: Given the equation, Find the center and radius of the circle. Sketch and indicate the center of the graph.
x^2+y^2-2x=15
x^2+6x+y^2+8=12
Solution: Rewrite the equation in standard form by completing the square in x and y. We can determine the center and the radius using the standard form of the equation.
x^2+y^2-2x=15
x^2+2x+1+y^2=15+1
(x+1)^2+y^2=16
Center (1,0) , r=4

x^2+6x+y^2+8y=12
x^2+6x+9+y^2+8y+16=12+9+16
(x+3)^2+(y+4)^2=37
Center (3,4), r=?37