Question

find an equation of the circle that has center (-2,5) and passes through (2,-2)

Answer 1

If (x0,y0) is the centre of the circle with radius r, then the equation would be
(x-x0)^2+(y-y0)^2=r^2
Here you know (x0,y0) but need r.
So substitute (x,y)=(2,-2), and (x0,y0)=(-2,5) in the above equation of the circle to find r^2 to complete the answer.

Answer 2

so 65?

Answer 3

Yes, 65 equals r^2.
So can you now show the complete equation of the circle?

Answer 4

r^2=65?

Answer 5

Use
" the equation would be (x-x0)^2+(y-y0)^2=r^2"
and substitute x0, y0 and r^2 using known values.

Answer 6

how would that look/

Answer 7

this is the part I have trouble with

Answer 8

The general equation of a circle with centre (x0,y0) and radius r is given by:
(x-x0)^2+(y-y0)^2=r^2

Since you are given the coordinates of the centre (x0,y0), and you have calculated r^2, you answer to the question will be an equation in x and y, but using numeric values for x0, y0 and r^2.

Answer 9

so its 2(-2)^2+-2(5)^2=65^2

Answer 10

2(-2(0)^2+-2(0)^2=65

Answer 11

like that?

Answer 12

Since we don't know x and y, they stay in the equation. We only substitute x0, y0 and r^2.
Can you give it another try?

Answer 13

x-(-2)^2+y-(5)^2=65

Answer 14

It is almost good, except that the parentheses are not perfect.
It should read:
(x-(-2))^2+(y-5)^2=65
which simplifies to
(x+2)^2+(y-5)^2=65

Answer 15

thank you!

Answer 16

yw! :)