Answer:
center = [tex](-4,8)[/tex]
Radius = 7 units
Step-by-step explanation:
Given: Equation of circle is [tex]x^2+y^2+8x-16y+31=0[/tex]
To find: Radius and center of the circle
Solution:
Equation of circle is [tex](x-a)^2+(y-b)^2=r^2[/tex]
Here, [tex](a,b)[/tex] is the center and r is the radius.
[tex]x^2+y^2+8x-16y+31=0\\\left [ x^2+2(4)x+4^2 \right ]+\left [ y^2-2(8)y+8^2 \right ]+31=4^2+8^2[/tex]
Use formula [tex](u+v)^2=u^2+v^2+2uv[/tex]
[tex](x+4)^2+(y-8)^2=16+64-31\\(x+4)^2+(y-8)^2=49=7^2[/tex]
On comparing this equation with equation of circle,
center = [tex](-4,8)[/tex]
Radius = 7 units
