Question

Please help!
Convert to standard form:

Answer 1

\[x ^{2}-6x+y ^{2}-12y+41=0\]

Answer 2

i got aaaand im stuck

Answer 3

the standard form of circle is (x-h)^2+(y-k)^2=r^2

Answer 4

@ganeshie8

Answer 5

i dont know where i was going with the factoring

Answer 6

good try, but looks like there is a mistake

Answer 7

\(\large x ^{2}-6x+y ^{2}-12y+41=0\)

step1 : group the variable terms separately
\(\large (x ^{2}-6x)+(y ^{2}-12y) = -41\)

Answer 8

step2 : complete the square for the stuff inside parenthesis

Answer 9

6/2 = 3
so add 3^2 both sides

12/2 = 6
so add 6^2 both sides

Answer 10

oh is it just (x-3)^2+(y-6)^2

Answer 11

Yes !

Answer 12

oh and i add the 9 and 36 to -41?

Answer 13

Exactly !

Answer 14

to get 4

Answer 15

yess

Answer 16

center is at (3,6) so i can get the h,k and put it in the standard form for a circle, but how do i find r? or is the r the 4?

Answer 17

again, what is the standard form of circle ?

Answer 18

\[(x-h)^{2}+(y-k)^{^{2}}=r ^{2}\]

Answer 19

so

Answer 20

and you got :
\[(x-3)^2 + (y - 6)^2 = 4\]

Answer 21

\(\implies r^2 = 4\)

\(r = ?\)

Answer 22

\[(x-3)^{2}+(y-6)^{2}=r ^{2}\]

Answer 23

so r=2

Answer 24

Correct !!

Answer 25

yay, thanks!!!

Answer 26

yw :) good job !