Question

Pre calculus help! Find the standard equation, center and radius.
x^2+y^2+6y+9=1/16

Answer 1

focus on this part:
\[y^2 + 6y + 9\]
can you write that in the form :
\[(y - a)^2\]

Answer 2

Im not really sure how to do that.

Answer 3

could it be (y-3)^2?

Answer 4

\[(y-3)^2=y^2-6y+9\]
try
\[(y+3)^2\]

Answer 5

y+6?

Answer 6

\[(x-a)^2=x^2-2ax+a^2\]
this is general formula.
\[(x+3)^2=(x-(-3))^2+x^2-2x(-3)+(-3)^2=x^2+6x+9\]

Answer 7

So, all i'd do is solve it?

Answer 8

Have you found the standard equation?

Answer 9

No, I haven't.

Answer 10

@.Sam. this is not linear equation

Answer 11

\[(x – h)^2 + (y – k)^2 = r^2?\]

Answer 12

yup, that's it

Answer 13

Teach her how to get to it

Answer 14

I'm trying. Now that we found
\[(y+3)^2=y^2+6y+9\],
Can you substitute this into your original equation?

Answer 15

Substitute how?

Answer 16

since your original equation is:
\[x^2+y^2+6y+9=\frac{1}{16}\]
and we've figured:
\[y^2+6y+9=(y+3)^2\]
we can now write down:
\[x^2+(y+3)^2=\frac{1}{16}\]

Answer 17

the standard formula has the following form:
\[(x-a)^2+(y-b)^2=r^2\]
wher (a,b) is the center and r is the radius

Answer 18

so in our case, what is the center of a circle?

Answer 19

3 and x?

Answer 20

we can write down our equation in slightly different form:
\[(x-0)^2+(y-(-3))^2=\frac{1}{16}\]

Answer 21

can you see it now?

Answer 22

0 and 3?

Answer 23

0 and -3

Answer 24

if we had (y-3)^2 it would be 3

Answer 25

but we've got (y+3)^2 = (y-(-3))^2

Answer 26

So, how do we find the center?

Answer 27

Radius! Sorry.

Answer 28

we can see that \[r^2=\frac{1}{16}\]

Answer 29

thus\[r=\sqrt{\frac{1}{16}}\]

Answer 30

so \[r = \frac{ \sqrt{1} }{ \sqrt{16}}=\frac{ 1 }{ 4 }\]

Answer 31

I think i understand it.

Answer 32

glad to hear it :)

Answer 33

Thank you so much!

Answer 34

@Fifciol
It would be nice for her to start with the standard equation, so that she appreciates every step that you're showing her to do:
As @.sam. said, the objective is:
\(\large (x – h)^2 + (y – k)^2 = r^2\)
It is much easier for a tutoree to follow you, and be motivated if he/she knows what the objective to start with. It also encourages him/her to try working out the problem, and to solve similar problems later on, which is the purpose of your lesson.

Answer 35

right, maybe I'm not a good teacher :(