ANSWER
[tex]\begin{gathered} Center:(4,-5) \\ Radius:5 \end{gathered}[/tex]EXPLANATION
We want to find the center and radius of the circle with the given equation:
[tex]x^2+y^2-8x+10y=-16[/tex]To do this, we have to write the equation in standard form:
[tex](x-h)^2+(x-k)^2=r^2[/tex]where (h, k) = center
r = radius
To do this, complete the square for both x and y variables of the equation.
Let us do that now:
[tex]\begin{gathered} x^2-8x+y^2+10y=-16 \\ x^2-8x+(-\frac{8}{2})^2+y^2+10y+(\frac{10}{2})^2=-16+(-\frac{8}{2})^2+(\frac{10}{2})^2 \\ x^2-8x+16+y^2+10y+25=-16+16+25 \\ x^2-8x+16+y^2+10y+25=25 \end{gathered}[/tex]Finally, factorize both the x and y variables:
[tex]\begin{gathered} x^2-4x-4x+16+y^2+5y+5y+25=25 \\ x(x-4)-4(x-4)+y(y+5)+5(y+5)=25 \\ (x-4)(x-4)+(y+5)(y+5)=25 \\ (x-4)^2+(y+5)^2=5^2 \end{gathered}[/tex]Therefore, the center and radius of the circle are:
[tex]\begin{gathered} Center:(4,-5) \\ Radius:5 \end{gathered}[/tex]That is the answer.
