Answer:
Centre = (1, 3)
radius = 8units
Explanation:
The standard equaton of a circle is expressed as;
[tex]x^2+y^2+2gx+2fy+C\text{ = 0}[/tex]The radius of the circle is expressed as;
[tex]r=\sqrt[]{g^2+f^2-C^{}}[/tex]The centre is at C(-g, -f)
Given the expression;
[tex]x^2+y^2-2x-6y-54=0[/tex]Get the centre of the circle.
Compare both equations
[tex]\begin{gathered} 2gx=-2x \\ g\text{ = -1} \end{gathered}[/tex]similarly;
[tex]\begin{gathered} 2fy=-6y \\ 2f=-6 \\ f=-\frac{6}{2} \\ f=-3 \end{gathered}[/tex]The centre will be located at C(-(-1), -(-3)) = C(1, 3)
Next is to get the radius
Recall;
[tex]\begin{gathered} r=\sqrt[]{g^2+f^2-C} \\ r=\sqrt[]{(-1)^2+(-3)^2-(-54)} \\ r=\sqrt[]{1+9+54} \\ r=\sqrt[]{64} \\ r=8\text{units} \end{gathered}[/tex]Hence the radius is 8units