1. ## circles- geometry-coordinates

a circle passes through the points (0,-3) and (0,3) and has radius 5

find equations of possible solutions (two possible)

realllyyy confused even though it's probably quite simple, even unsure where to begin

thanks

2. Originally Posted by lra11
a circle passes through the points (0,-3) and (0,3) and has radius 5

find equations of possible solutions (two possible)

realllyyy confused even though it's probably quite simple, even unsure where to begin

thanks
Call the center of the circle (h, k). Then the circle equation is
$(x - h)^2 + (y - k)^2 = r^2$

Thus
$(0 - h)^2 + (-3 - k)^2 = 5^2$
and
$(0 - h)^2 + (3 - k)^2 = 5^2$

Solve for h and k.

-Dan

3. Think about (0,4) & (0,-4) as possible centers.

4. A circle has a center (h,k) and a radius r in the formula
(x-h)^2 + (y-k)^2=r^2. You will end up with two equations when plugging in your information:

1) (0-h)^2 + (-3-k)^2 = 5^2
2) (0-h)^2 + (+3-k)^2 = 5^2

Try to solve for h and k from using the system of two quadratics equations...

5. ## (0,4)

is this because 5+3=8 / 2 = 4 or am i on the wrong lines completely

6. Originally Posted by lra11
is this because 5+3=8 / 2 = 4 or am i on the wrong lines completely
Pet peeve warning! Danger Will Robinson! Danger! Danger!

How can 5 + 3 = 8/2 = 4?

Look at the outsides of this: 5 + 3 = 4 This is obviously not true! You need to write the expressions correctly, so this needs to go on two lines:
5 + 3 = 8
8/2 = 4

Okay, pet peeve rant over.

-Dan

7. h^2 + k^2 + 6k + 9 = 25
h^2 + k^2 - 6k + 9 = 25

so

h^2 + k^2 + 6k + 9 = 25
-h^2 + -k^2 + 6k - 9 = -25

12k = 0
k = 0

Therefore,
h^2 + 0^2 + 6(0) + 9 = 25
h^2 + 0^2 - 6(0) + 9 = 25

so

h^2 + 9 = 25

h^2 = 25-9

h^2=16

h= +-4

Centers are (4,0) and (-4,0)

8. Come on! All that was needed is to note that you have a 3,4, 5 triangle.
All that other stuff is needless.

9. ## thanks

thanks everyone for the help, it seems so simple now.

10. Needless? Sometimes shortcuts aren't the best way to get to the answer. The longer method works whether the triangle is 3-4-5 or say...4-5-sqrt 41 or whatever.

I'm sure they will learn and benefit from BOTH approaches...man...chill out...

11. 3-4-5 triangles are geometry.

Solving for variables is coordinate geometry

12. Yes, and given that this question was in the context of "coordinate geometry", my approach shouldn't have been an issue with Plato.

Sheesh...

13. Originally Posted by mathdeity
Yes, and given that this question was in the context of "coordinate geometry", my approach shouldn't have been an issue with Plato.

Sheesh...
Well, whatever works, works, I guess. You could approach this problem from many areas of math, just like the 100+ proofs of the pythagorean theorem.