Question

In the xy-plane, the graph of x2 − 4x + y2 − 6y = −8 is a circle. What is the radius of the circle?

Answer 1

The center-radius form of the circle equation

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have:

$x^2-4x+y^2-6y=-8$

Use

$(a-b)^2=a^2-2ab+b^2\qquad(*)$

$x^2-2(x)(2)+y^2-2(y)(3)=-8\qquad\text{add}\ 2^2\ \text{and}\ 3^2\ \text{to both sides}\\\\\underbrace{x^2-2(x)(2)+2^2}_{(*)}+\underbrace{y^2-2(y)(3)+3^2}_{(*)}=2^2+3^2-8\\\\(x-2)^2+(y-3)^2=4+9-8\\\\(x-2)^2+(y-3)^2=5\\\\(x-2)^2+(y-3)^2=(\sqrt5)^2\\\\Answer:\\\\\boxed{center:(2,\ 3)}\\\\\boxed{radius:r=\sqrt5}$