Answer:
[tex](x-3)^2 + y^2 = 68[/tex]
Step-by-step explanation:
The vertex form of the equation of a circle is [tex](x-h)^2 + (y-k)^2 = r^2[/tex] where (h,k) is the center of the circle and r is the radius. This circle has center (3,0). Find the radius using the distance formula with (3,0) and (1,8).
[tex]d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\d = \sqrt{(3-1)^2 + (0-8)^2}\\d = \sqrt{(2)^2 + (-8)^2} \\d = \sqrt{4 + 64} =\sqrt{68}[/tex]
Substitute h = 3, k = 0 and r = √68 into the vertex form. Then simplify for the equation.
[tex](x-h)^2 + (y-k)^2 = r^2\\(x-3)^2 + (y-0)^2 = \sqrt{68}^2\\ (x-3)^2 + y^2 = 68[/tex]
