Question

X2+ y2 - 8x - 12y + 27 = 0 Center: Radius:

Answer 1

Answer:

Give George brainliest!

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation:

You have:

$x^2+y^2-8x+27=0$

Since the variable 'x' appears twice, I am going to choose to group together the terms as to create a 2nd degree trinomial:

$(x^2-8x+27)+y^2=0$

This allows us to complete the square and factor the trinomial:

$(x^2-8x+16-16+27)+y^2=0$

$(x^2-8x+16)+11+y^2=0$

$(x-4)^2+y^2+11=0$

$(x-4)^2+(y-0)^2=-11$

This is not a valid circle equation in the real plane. It has a complex radius of:

$r=i\sqrt{11}$

And is centered at the point:

$(4,0)$