Question

Please help me :(...The circle given by x^2+y^2-6y-12=0 can be written in standard form like this:x^2(y-k)^2=21...what is the value of k in this equation?

Answer 1

x^2+(y-3)^2=21

Answer 2

could you explain more please

Answer 3

k=3

Answer 4

you must complete the square:

\[y^{2} + by = (y + b/2)^{2} - (b/2)^{2}\]


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

Answer 5

so it is 3 ?

Answer 6

so the end equation is \[x^{2} + (y-3)^{2} - 9 - 12 = 0\]
x^{2} + (y-3)^{2} = 21

Answer 7

yes k = 3

Answer 8

thanks guys,big help !

Answer 9

glad i could help :) . remember and practice square completion. you'll use it later

Answer 10

i think the equation of a circle is \[x ^{2}+y ^{2}=a ^{2}\]