I need help solving the equation for my revision class

Question

I need to solve the radius and the center point. $$ x^2+y^2=6x-2y-8 $$ Under this is my attempt of solving it $$ x^2-6x+y^2+2y=-8 $$ $$ x^2-6x+9+y^2\ +2x+1=-8+1+9 $$ $$ \left(x+3\right)^2+\left(y-1\right)^2=2^2 $$ On 2nd I added the integers to both sides

Answer 1

By solving you can get $$x^2+y^2=6x−2y−8$$

$$(x^2-6x+9-9) + (y^2+2y+1-1) =-8$$

$$(x-3)^2 +(y+1)^2 = 2.$$

Answer 2

Close but no cigars :-)

At the right side you have $=2$, which does not equal $=2^2$ but $=(\sqrt{2})^2$

:-)

Answer 3

$x^2+y^2=6x-2y-8$

$\Rightarrow x^2-6x+y^2+2y=-8$

$\Rightarrow x^2-6x+9+y^2+2y+1=-8+1+9$

(You have written wrongly $2x$ instead of $2y$)

$\Rightarrow (x^2-6x+9)+(y^2+2y+1)=2$

$\Rightarrow \left(x-3\right)^2+\left(y+1\right)^2={(\sqrt2)}^2$

(You have done wrong calculation here)

We know that,the equation of a circle with center at $(a,b)$ and with radius $r$ is, ${(x-a)}^2+ {(y-b)}^2= r^2$.

So comparing with above equation we get that, the center is $(3,-1)$ and radius is $\sqrt 2$.