Question

X2+y2-10x-16y+53=0 what is the center and 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:

$x^2+y^2-10x-16y+53=0$

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

$x^2-10x+y^2-16y+53=0\qquad\text{subtract 53 from both sides}\\\\x^2-2(x)(5)+y^2-2(y)(8)=-53\qquad\text{add}\ 5^2\ \text{and}\ 8^2\ \text{to both sides}\\\\\underbrace{x^2-2(x)(5)+5^2}_{(*)}+\underbrace{y^2-2(y)(8)+8^2}_{(*)}=5^2+8^2-53\\\\(x-5)^2+(y-8)^2=25+64-53\\\\(x-5)^2+(y-8)^2=36\\\\(x-5)^2+(y-8)^2=6^2\\\\Answer:\\\\\boxed{center:(5,\ 8)}\\\\\boxed{radius:r=6}$