circle whose center is at the origin and whose radius is
Which of the following points lies on the circle whose center is at the origin and whose radius is 10? I'm not good with this tho :3
Given center and radius, can you write the equation of the circle ?
Wait this?\[r=\sqrt{(x-0)^2 +(y-0)}\]
Yes, plugin r = 10 too
Wait so i have to rewrite the fomula again?
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Also it seems you've forgot to square the y part
Wait this?\[r=\sqrt{(x-0)^2 +(y-0)^{\color{red}{2}}}\]
Oh yea 10=\(\sqrt{(x-0)^2 +(y-0)^{\color{}{2}}} \)
square both sides
**10=\(\sqrt{(x-10)^2 +(y-10)^{\color{}{2}}}\) liek this? sorry i'mnot good with this
No. (0,0) is the center. Your earlier equations are correct
10=\(\sqrt{(x-0)^2 +(y-0)^{\color{}{2}}} \) squaring both sides gives 100 = \(x^2 + y^2\)
Any point on the circle must satisfy above equation
plugin each of the given options and see which one satisfies
Okay so this way x2+y2=100 \[\sqrt{10}^2 +\sqrt{10}^2\] Like this?
what are your options ?
Uhh let me see (-6, 4) (\(\sqrt{10},\sqrt{10}\) ) (6, -8)
Lets test the first option (-6, 4) plugin x = -6, y = 4 in the equation 100 = x^2 + y^2
100 = (-6)^2 + 4^2 100 = 36 + 16 100 = 52 which is not true so the point in the first option doesn't lie on the given circle
try the second option
Okay let me see I got 20 as 100
which is clearly false try third option
Oh i was thinking it's false :3 mkay 3rd one \(6^2+(-8)^2 is 100\)
So ?
Oh probably c, but i'm not sure.. let me check that agiain
O.m.g that's right
Yes
You could also solve it by graphing
Simply graph the equation of the circle and the given points
YAAAAAAS!1!!!!!! OKay Ummm should i try it?
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