Question

HELP PLEASE!!! Write the standard form of the equation for the circle that passes through the points (-9, -16), (-9, 32), and (22, 15). Then identify the center and radius. MULTIPLE CHOICE.

Answer 1

Use this: The equation of a circle : (x-a)2 + (y-b)2 = r2
(a,b) = center
r = radius

Answer 2

a. (x-3)^2 + (y+9)^2 = 25
(-2, 8), 25

b. (x+2)^2 + (y-8)^2 =625
(-2, 8), 25

c. (x+2)^2 + (y-8)^2 = 625
(-3, 9), 5

d. (x-3)^2 + (y+9)^2 = 25
(-3, 9), 5

Answer 3

@Blank  @navk

Answer 4

@Tom_Boy_Rebel how do you plug in the equation?

Answer 5

Plug in the points in the choices to check the answers. The equation which is satisfied by all the three given points is the right one.

Answer 6

Answer is:

b. (x+2)^2 + (y-8)^2 =625
(-2, 8), 25

Answer 7

Ok I get it. Thank you @navk and @Blank 

Answer 8

Or you can also find the equation from the points which is quite a long method. For that you need to use the general equation x^2 + y^2 + 2gx + 2fy + c = 0

Answer 9

Thank you :)

Answer 10

ǝɯoɔlǝʍ ǝɹ,noʎ :)

Answer 11

@Blank  can you help me with one more?

Answer 12

I'll try but I'm really busy.

Answer 13

Write the standard form of the equation for the circle that passes through the points (30, -2), (-1, -19), and (-18, 12). Then identify the center and radius.

Answer 14

a. (x+6)^2 + (y-5)^2 = 625
(6, 5), 25

b. (x-6)^2 +(y-5)^2 = 625
(6, 5), 25

c. (x+5)^2 - (y+6)^2 = 650
(3, 5), 15

d. (x+5)^2 + (y+6)^2 = 625
(6, 2), 25

Answer 15

@Blank 

Answer 16

Just a second...

Answer 17

b. (x-6)^2 +(y-5)^2 = 625
(6, 5), 25

Answer 18

Thank you so much! @Blank 

Answer 19

ǝɯoɔlǝʍ ǝɹ,noʎ :)