Question

Find the radius of the circle with these three points: (-4,9), (2,9), and (-5,8) and center: (-1,5).

Answer 1

The center tells you alot: That gives you this without any work:
\[(x+1)^2+(y-5)^2=r^2\].
From here you can plug in x and y and solve directly for r.

Answer 2

which x and y

Answer 3

Any of them :) They all have to satisfy the equation. So they should all give the same r.

Answer 4

Do you have to foil both sides?

Answer 5

You don't have to. For example try the point (-4,9).
You get:
\[((-4)+1)^2+(9-5)^2=r^2\]
\[(-3)^2+(4)^2=r^2\]
\[9+16=r^2\]
\[25=r^2\]
r=5

Answer 6

would it be -5

Answer 7

The radius is always defined as a positive number. But 5 is correct :)

Answer 8

thanks for the help malevolence19 {;

Answer 9

Do you know how to solve Find the center of the circle defined by the equation
6x^2+6y^2+48x-36y-144=0

Answer 10

Yes. You complete the square.
\[6(x^2+8x)+6(y^2-6y)=144\]. (rewriting)
Then:
\[6(x^2+8x+16-16)+6(y^2-6y+9-9)=144\]
Then factoring:
\[6(x+4)^2+6(y-3)^2-16(6)-9(6)=144\]
\[6[(x+8)^2+(y-3)^2]=204\]
\[(x+8)^2+(y-3)^2=34\]

Answer 11

Err those last lines should say (x+4)

Answer 12

Wrong. So 144+54+96=294. Then dividing by 3 gives 49. So you should have:
(x+4)^2+(y-3)^2=49.

Sorry I just failed so hard >.<