Question

7. Graph the circle with center at (-3, 2), which also passes through the point (0, -2). Label the center and at least four points on the circle. Write the equation of the circle. @jim_thompson5910

Answer 1

you have the center, but the radius is missing

Answer 2

to find the radius, we need to find the distance from (-3,2) to (0,-2)

Answer 3

how do we do this?

Answer 4

idk can you help me finding it

Answer 5

are you familiar with the distance formula

Answer 6

not really

Answer 7

the distance formula is this

d = sqrt((x2-x1)^2+(y2-y1)^2)

look familiar?

Answer 8

a little bit :)

Answer 9

ok great

Answer 10

The two generic points are (x1, y1) and (x2, y2)

Answer 11

so what do we plug in?

Answer 12

In this case, the two points are (-3, 2) and (0, -2)

Answer 13

Which means...

(x1, y1) = (-3, 2) ----> x1=-3 and y1= 2

(x2, y2) = (0, -2) ----> x2=0 and y2= -2


So in short,

x1=-3
y1= 2
x2=0
y2= -2

Answer 14

Plug these values in to get


d = sqrt((x2-x1)^2+(y2-y1)^2)


d = sqrt((0-(-3))^2+(-2-2)^2)


d = sqrt((0+3)^2+( -2- 2)^2)


d = sqrt((3)^2+(-4)^2)


And I'll let you finish

Answer 15

(x2-x1)^2+(y2-y1)^2)
0-(-3)^2+-2-2^2

Answer 16

good (sorta)

Answer 17

you keep going to get


d = sqrt((x2-x1)^2+(y2-y1)^2)


d = sqrt((0-(-3))^2+(-2-2)^2)


d = sqrt((0+3)^2+( -2- 2)^2)


d = sqrt((3)^2+(-4)^2)


d = sqrt(9+16)


d = ???

Answer 18

sqrt 25
sqrt5 @jim_thompson5910

Answer 19

once you take the square root of 25, you will get 5 (and the square root will go away)

Answer 20

in other words \[\Large \sqrt{25} = 5\]

Answer 21

so d = 5

Answer 22

which means that the distance from (-3, 2) to (0, -2) is 5 units

Answer 23

making the radius to be 5 units

so r = 5

Answer 24

so whats the center?

Answer 25

it's given above

Answer 26

at the very top

Answer 27

-3,2?

Answer 28

(-3,2) yes

Answer 29

use that with the radius to find the equation