Question

A certain circle can be represented by the following equation. x2 + y2 + 10x + 12y + 25 = 0. What is the center of this circle? What is the radius of this circle? Please help!

Answer 1

Let's try to complete the squares.

The x-part starts with $x^2+10x$, which is the beginning of $x^2+10x+25=(x+5)^2$. So, we'll think of $x^2+10x$ as $(x+5)^2-25$

Similarly, we have that

$y^2+12y = (y+6)^2-36$

So, the equation becomes

$x^2 + y^2 + 10x + 12y + 25 = 0 \iff (x+5)^2-25 + (y+6)^2-36+25=0 \iff (x+5)^2+ (y+6)^2-36=0 \iff (x+5)^2+ (y+6)^2=36$

Now we have writte the equation of the circle in the form

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

When the equation is in this form, everything is more simple: the center is $(k,h)$ and the radius is $r$.

Answer 2

Answer:

Center// (-5,-6)

Radius// 6