Answer: [tex]r=3[/tex]
Step-by-step explanation:
The equation of the circle in center-radius form is:
[tex](x-h)^2+(y-k)^2=r^2[/tex]
Where the point [tex](h,k)[/tex] is the center of the circle and "r" is the radius.
Subtract 56 from both sides of the equation:
[tex]x^2+y^2+8x-14y+56-56=0-56\\x^2+y^2+8x-14y=-56[/tex]
Make two groups for variable "x" and variable "y":
[tex](x^2+8x)+(y^2-14y)=-56[/tex]
Complete the square:
Add [tex](\frac{8}{2})^2=4^2[/tex] inside the parentheses of "x".
Add [tex](\frac{14}{2})^2=7^2[/tex] inside the parentheses of "y".
Add [tex]4^2[/tex] and [tex]7^2[/tex] to the right side of the equation.
Then:
[tex](x^2+8x+4^2)+(y^2-14y+7^2)=-56+4^2+7^2\\\\(x^2+8x+4^2)+(y^2-14y+7^2)=9[/tex]
Rewriting, you get that the equation of the circle in center-radius form is:
[tex](x+4)^2+(y-7)^2=3^2[/tex]
You can observe that the radius of the circle is:
[tex]r=3[/tex]
