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 4Solution: Equation of the circle: (x-h)^2+(y-k)^2=r^2The 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^2r= (2)^2+(5)^2=29Therefore 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=12Solution: 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=15x^2+2x+1+y^2=15+1(x+1)^2+y^2=16Center (1,0) , r=4 x^2+6x+y^2+8y=12x^2+6x+9+y^2+8y+16=12+9+16(x+3)^2+(y+4)^2=37Center (3,4), r=?37