Question

please help trying to find the end points for the circle. (x-2)^2+(y-6)^2=16

Answer 1

you will never make it, a circle has no end points..

Answer 2

could you tell what you mean by endpoints ?

Answer 3

sorry i meant when x and y pass 0. x=0 and y =0 so (0, ) ( ,0)

Answer 4

plugin x=0 in the equation and solve y

Answer 5

(x-2)^2+(y-6)^2=16

(0-2)^2+(y-6)^2=16

Answer 6

can you solve y ?

Answer 7

so i just plug in y? i dont have to divide by 16?

Answer 8

do you have options ?

Answer 9

y=+-4.47?

Answer 10

nope, im getting y = 2.54, 9,46

Answer 11

and there are no x intercepts.. can you check the equation again ?

Answer 12

sorry i didnt type it (x-2)^2+(y-6)^2=16

Answer 13

could you take a screenshot of the complete question and attach ?

Answer 14

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