Question

The center and radius of the circle below is in the form (h,k) and r, respectively, find h+k+r.x^2+y^2+6x-2y-15=0

Answer 1

h + k + r = 3

Step-by-step explanation:
`\begin{gathered} \text{Given equation of circle:} \\ x^2+y^2+6x-2y-15=0​ \end{gathered}`

The equation is in general form, so we use the general form of the equation to find h,k, and r

`\begin{gathered} \text{General form of equation of cirlce:} \\ x^2+y^2+2gx+2fy+c=0​ \\ \text{where center (}-g,\text{ -f)} \\ \text{radius = }\sqrt[]{g^2+f^2-c} \end{gathered}`
`\begin{gathered} \text{comparing the given equation with the standard form of equation of circle:} \\ 2gx\text{ = 6x} \\ 2g\text{ = 6} \\ g\text{ = 6/2 = 3} \\ 2fy\text{ = -2y} \\ 2f\text{ = -2} \\ f\text{ = -2/2 = -1} \\ c\text{ = -15} \end{gathered}`
`\begin{gathered} r\text{= }\sqrt[]{3^2+(-1)^2-(-15)} \\ r\text{= }\sqrt[]{9+1+15} \\ r\text{ = }\sqrt[]{25}\text{ = 5} \end{gathered}`
`\begin{gathered} In\text{ vertex form, } \\ \text{center of circle =(h, k)} \\ In\text{ standard form,} \\ \text{center of circle =(-g,-f)} \\ \text{compare with the center of circle in the standard equation of circle}\colon \\ h\text{ = -g} \\ h\text{ = -(3) = -3} \\ k\text{ = -f} \\ k\text{ = -(-1) = 1} \\ \end{gathered}`
`\begin{gathered} h\text{ + k + r = -3 + 1 +5} \\ \text{ }h\text{ + k + r = -3 + 1 + 5} \\ h\text{ + k + r = 3} \end{gathered}`