[tex]\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{6}{ h},\stackrel{-3}{ k})\qquad \qquad radius=\stackrel{5}{ r}\\[2em] [x-6]^2+[y-(-3)]^2=5^2\implies (x-6)^2+(y+3)^2=25[/tex]
Answer:
Step-by-step explanation:
The x value for the center is part of the x^2 part of the equation.
The y value for the center is part of the y^2 part of the equation.
The radius is squared.
(x - 6)^2 + (y + 3)^2 = 5^2
Notice the sign change for the center. When you move horizontally the sign of the center changes sign in the circle's equation.
The graph is included to show you that.
