Given:
[tex]x^2+y^2+6x+8y-75=0[/tex]To find the circle equation in center and radius form:
Let us rewrite the equation as
[tex]\begin{gathered} x^2+y^2+6x+8y-75=0 \\ x^2+6x+y^2+8y-75=0 \\ (x^2+6x+3^2)-3^2+(y^2+8y+4^2)-4^2-75=0 \\ (x+3)^2-9+(y+4)^2-16-75=0 \\ (x+3)^2-9+(y+4)^2-16-75=0 \\ (x+3)^2+(y+4)^2-100=0 \\ (x+3)^2+(y+4)^2=100 \end{gathered}[/tex]Thus, the equation of the circle in center-radius form is,
[tex](x+3)^2+(y+4)^2=100[/tex]The general equation of the circle in center-radius form is,
[tex](x-h)^2_{}+(y-k)^2=r^2[/tex]Where (h, k) is the center and r is the radius.
So, comparing we get
The center is,
[tex](-3,-4)[/tex]The radius is,
[tex]r=10[/tex]The equation of the circle in center-radius form is,
[tex](x+3)^2+(y+4)^2=100[/tex]