Question

What is the center and radius of x2 = -y2 + 8y +119

Answer 1

Given

`x^2=-y^2+8y+119`

It is equivalent to

`\begin{gathered} \Rightarrow x^2+y^2-8y=119 \\ \Rightarrow(x-0)^2+y^2-8y=119 \end{gathered}`

Complete the square for the y variable, as shown below

`\begin{gathered} (y-a)^2=y^2-8y \\ \Rightarrow y^2-2ay+a^2=y^2-8y \\ \Rightarrow-2a=-8 \\ \Rightarrow a=4 \\ \Rightarrow(y-4)^2=y^2-8y+16 \end{gathered}`

Therefore,

`\begin{gathered} \Rightarrow(x-0)^2+y^2-8y+16=119+16 \\ \Rightarrow(x-0)^2+(y-4)^2=135 \end{gathered}`

Thus, the center of the circumference is (0,4) and its radius is equal to 3sqrt(15)

`\begin{gathered} r^2=135 \\ \Rightarrow r=\sqrt[]{3^2\cdot15} \\ \Rightarrow r=3\sqrt[]{15} \end{gathered}`