Question

Using the following equation, find the center and radius:

x2 −2x + y2 − 6y = 26

The center is located at (−1, −3), and the radius is 36.

The center is located at (1, 3), and the radius is 36.

The center is located at (−1, − 3), and the radius is 6.

The center is located at (1, 3), and the radius is 6.

Answer 1

Complete the square:

x^2-2x + ?

Answer 2

\[x^2 −2x + y^2 − 6y = 26\]\[x^2 −2x \color{red} { +1 } + y^2 − 6y\color{blue} { +9 } = 26\color{red} { +1 }\color{blue} { +9 }\]

Answer 3

Complete the square again:
y^-6y+ ?

Answer 4

I had a similar question on a recent test and didn't know how to answer it either.

Answer 5

woah

Answer 6

see what I am doing in my post ?

Answer 7

I see that you used +1 and +9 but don't really understand where you got the numbers from?

Answer 8

I add them to both sides, I am striving to get something like

(x+C)^2+(y+D)^2...

Answer 9

Ok i understand what you did there now, had to think about it

Answer 10

okay, so can you now write the equation of the circle ?

Answer 11

(x-h)^2 + (y-k)^2 = r^2
(x-(-1))^2 + (y+-3)=r^2
(x-1)^2 + (y-3)^2 = r^2

Or am i going in the completely wrong direction with this, sorry if i am .-.

Answer 12

the form is\[(x-\color{red} { h })^2+(y-\color{green} { k })^2=\color{blue} { r }^2\]

Answer 13

i wrote that for the first part but don't you fill it out with -1 and -3 and 1 and 3? like i said, sorry if im going in the wrong direction with this

Answer 14

split the equation into 2 halves\[x^2-2x+1~~~-->~~~(x-1)^2\]\[y^2-6y+9~~~-->~~~(y-3)^2\]\[26+1+9~~~-->~~~36~~~-->~~~6^2\]

Answer 15

Ohhh ok i think i get it now

Answer 16

but is it (-1)^2 and (-3)^2 or (1)^2 and (3)^2? don't you get the same result either way or am i thinking about that wrong also

Answer 17

it is (x-1)^2 + (y-3)^2=8^2

Answer 18

ok i see that now too, thanks so much man