Question

Identify the center and radius from the equation of the circle given below. x^2+y^2+121-20y=-10x

Answer 1

Answer:

Center: (-5,10)

Radius: 2

Step-by-step explanation:

The equation of the circle in center-radius form is:

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

Where the point (h,k)  is the center of the circle and "r" is the radius.

Subtract 121 from both sides of the equation:

$x^2+y^2+121-20y-121=-10x-121\\x^2+y^2-20y=-10x-121$

Add 10x to both sides:

$x^2+y^2-20y+10x=-10x-121+10x\\x^2+y^2-20y+10x=-121$

Make two groups for variable "x" and variable "y":

$(x^2+10x)+(y^2-20y)=-121$

Complete the square:

Add $(\frac{10}{2})^2=5^2$ inside the parentheses of "x".

Add  $(\frac{20}{2})^2=10^2$  inside the parentheses of "y".

Add $5^2$ and $10^2$ to the right side of the equation.

Then:

$(x^2+10x+5^2)+(y^2-20y+10^2)=-121+5^2+10^2\\(x^2+10x+5^2)+(y^2-20y+10^2)=4$

Rewriting, you get that the equation of the circle in center-radius form is:

 $(x+5)^2+(y-10)^2=2^2$

You can observe that the radius of the circle is:

$r=2$

And the center is:

$(h,k)=(-5,10)$

Answer 2

Answer:

Step-by-step explanation:

x²+y²+121-20y=-10x

(x²+10x)+(y²-20y)+121=0

(x²+10x+25)-25+(y²-20y+100)-100+121=0

(x+5)² + (y-10)²= 2²

the center is : A(-5;10)  and radius : r = 2