Circles 2

can anyone help answers these?

1) $r=5units$

Let's see why:

First rearrange the terms to make it easier to work with:

$x^2-12x+y^2+6y=-20$

I have grouped the x- and y-containing terms together. Now, we must use a method called completing the square. In a quadratic term, the equation is ax^2+bx+c=0. In the above equation, however, there is no c-term. To find the missing c-term, do (b/2)^2.

$x^2-12x+(\frac{b_1}{2})^2+y^2+6y+(\frac{b_2}{2})^2=-20+(\frac{b_1}{2})^2+(\frac{b_2}{2})^2$  Remember, whatever you do to one side, you must do to the other.

$x^2-12x+(\frac{-12}{2})^2+y^2+6y+(\frac{6}{2})^2=-20+(\frac{-12}{2})^2+(\frac{6}{2})^2$ Substituting the values 

$x^2-12x+36+y^2+6y+9=-20+36+9$

$x^2-12x+36+y^2+6y+9=25$

Okay, we have manipulating the equation so that the number on the right side of the equation is the r^2, You need not go further with the simplification of this problem as it is only asking for the radius length. We have enough information for this

$r^2=25$

$r=5units$Of course, include units in the final answer.

2) $C=(1,3).$

Here's why:

Just like the previous problem, I will rearrange the terms, so the equation ios easier to handle:

$x^2-2x+y^2-6y=-1$

We will use the same process as the previous problem: 

$x^2-2x+(\frac{b_1}{2})^2+y^2-6y+(\frac{b_1}{2})^2=-1+(\frac{b_1}{2})^2+(\frac{b_2}{2})^2$

$x^2-2x+(\frac{-2}{2})^2+y^2-6y+(\frac{-6}{2})^2=-1+(\frac{-2}{2})^2+(\frac{-6}{2})^2$

$x^2-2x+1+y^2-6y+9=-1+1+9$

$x^2-2x+1+y^2-6y+9=9$

In this case, the problem is asking for the center of the circle. The standard form for a circle is (x – h)² + (y – k)² = r²

, where (h,k) is the center of the circle. However, we still need to manipulate the equation further. :Luckily, x^2+2x+1 and y^2-6y+9 are both perfect square trinomials, which means that they can be represented as (x+(b/2))^2 and (y+(b/2))^2.

$(x+\frac{b_1}{2})^2+(y+\frac{b_2}{2})^2=9$The number that goes for b is always the b coefficient in quadratics. 

$(x+\frac{-2}{2})^2+(y+\frac{-6}{2})^2=9$Substituting

$(x-1)^2+(y-3)^2=9$Finally, after all this work, we have gotten to the standard equation of a circle. 

$-1=-h$,so $h=1$.

$-3=-k$,so $k=3$.

The center of a cricle is always $(h,k)$,so the center, or C, is (1,3).