Question

X2+y2-10x-16y+53=0 what is the center and radius of the circle ?

Answer 1

Center: (5,8)

Radius: 6

Explanation:

Answer 2

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}`