Question

Please help me... I'm extremely stressed over this.
Which of these points lies on a circle centered at A(3, 3) and passing through B(6, 5)?
C(1, 6)
D(6, 0)
E(0, 3)
F(3, -1)
G(3, 6)

Answer 1

Will medal, by the way.

Answer 2

Use the circumference formula

Answer 3

... I may have forgotten how to do that...

Answer 4

Look it up and then place the y and x values they gave you on the problem (sorry I'm on the phone so idk how to do the square root symbol)

Answer 5

I think it's (3,6), but I'm not sure... I'm always bad with Geometry
This is my second year taking it and if I don't pass, I won't graduate this year... :/

Answer 6

I tried Graphing it with Paint, but it's not working very well

Answer 7

It would have helped if they gave me a picture, but unfortunately they didn't..

Answer 8

https://answers.yahoo.com//index?qid=20150317101400AAtwJcJ
There is a picture of the circle just find the points :)

Answer 9

The link unfortunately didn't work

Answer 10

@daniellelovee Please look at this http://openstudy.com/study#/updates/5752e8cee4b0c82eef133452

Answer 11

Rebecca I can't look at that I'm currently using my phone and I cant click on the link

Answer 12

Try using the website mathisfun and plug in the center :) since the link didn't work
I'm trying to find a way to draw trough the phone app but I can't find it

Answer 13

Find the distance betweens points 6,5 and 3,3 using the distance formula. You now have your radius. Now do the same thing with each of the answer choices until you get the same value as the radius.

Answer 14

what exactly is the distance formula..?

Answer 15

Well, you see,
the standard equation of circle is \[(x-h)^{2} + (y-k)^{2} = a\]
, where h and k are the coordinates of the centre (i.e., (h,k))
(x,y) is any point on the circle
a is the radius.
We, find the radius by distance formula
\[\sqrt{(x1-x2)^{2} + (y1-y2)^{2}}\]
on solving you find,
a= \[\sqrt{13}\]
so, \[x^{2}\] = 13

So, 13= \[(x-3)^{2} + (y-3)^{2}\]

Now, solve it

Let's try to do it wittily now,
13= 4+9, where 4 and 9 are both perfect squares
So, you see, we can put

\[(x-3)^{2} = 4\]

or, x-3 = +-2= -2 (let's assume it's -ve)
or, x=3-2= 1

and, y-3 =+3, or, y-3=3 (again let's assume +ve)
so, y=3+3=6

so your answer is (1,6)

you can obtain answers by putting the options on 'em.
See, if there are other answers too, @Kittyloli

Answer 16

And I checked out, there's no other answers that satisfy the equation from those options.
I hope you find this helpful @Kittyloli