𝑥2+𝑦2−12𝑥+12𝑦+68=0 x 2 + y 2 − 12 x + 12 y + 68 = 0 is the equation of a circle with center (ℎ,𝑘) ( h , k ) and radius 𝑟 r for: h= and k= and r =

Answer 1

x² + y² - 12x + 12y + 68 = 0

Now the general formula for a circle will look like this

(x-h)²+(y-k)²=r²

where r will be the radius of the circle, h will be the x coordinate of the center of the circle and k will be the y coordinate of the center of the circle.

How do we find these with that crazy equation we started with ?

if we look at x² - 12x, what we need to do is "complete the square",; in other words, we find out what we would have to add to x² - 12x such that we can get a square expression (x-h)². Well

(x-h)² = x² - 2hx + h²

how do we know what h is? well from x² - 12x, -2hx would have to match -12x so h would have to be 6.

Now (x - 6)² = x² -2*6*x+(6*6) = x² - 12x + 36.so we can rewrite

x² +y²- 12x + 12y + 68 = 0 as

x² - 12x+36 + y²+ 12y + 68 -36 =0 (notice how i "took" the 36 away from the 68 term.

(x-6)² + y²+ 12y + 32 = 0

NOW, we complete the square for the y terms:

we need to add something to y²+ 12y to make it a complete square (y-k)²

now (y-k)² = y² -2ky + k²

so -2ky has to equal +12y, so k has to equal -6

Now (y-(-6)) = y+6

and (y+6)² = y² -12y + (-6)² = y² -12y + 36

Now taking the revised equation

(x-6)² + y²+ 12y + 32 = 0

(x-6)² + y²+ 12y + 36 + 32 - 36 =0

(x-6)² + (y+6)² -4 = 0

(x-6)² + (y+6)² = 4

Now, we recognize that 4 is equal to the square of the radius from the general equation of a circle, so

r = 2 and the center of the circle would be located at (6, -6) so h = 6 and k = - 6

Answer 2

First, you want to factor the initial equation so that it is of the form (x-h)^2 + (y-k)^2 =r^2. To do this, you add 4 to both sides of the equation. Then the left side is x^2-12x+y^2+12y+72, so we can separate the +72 into two +36's so it becomes x^2-12x+36+y^2+12y+36=4. Then, we can factor both of these binomials of x and of y into the form (x-6)^2+(y+6)^2=4. Since 4=2^2, we can replace it by 2^2, so we have our equation in the desired form, with a circle centered at (6,-6) with radius 2.

Answer 3

𝑥2+𝑦2−12𝑥+12𝑦+68=0

Complete the square

x² - 12x + 36 + y² + 12y + 36 = -68 + 36 + 36

(x-6)² + (y+ 6)² = 4

h= -6 k = 6 r = 4