What is the equation of a circle that passes through (2, 0) and has its center at M(4, 0)?

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What is the equation of a circle that passes through (2, 0) and has its center at M(4, 0)?

Mathematics
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please help!!!!
standard form of circle: (x - h)^2 + (y - k)^2 = r^2center is (h , k) and radius is rfor your circle, the center (h , k) is (1 , 2)so it will look like: (x - 1)^2 + (y - 2)^2 = r^2to find the radius squared, plug in the (x,y) point you know is on the circle -- (3 , 4)(3 - 1)^2 + (4 - 2)^2 = r^22^2 + 2^2 = 8 = r^2
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???
Ok, so we know the standard form of a circle is (x - h) ² + (y - k) ² = r ² and the center is (h , k) and radius is r. Does this make sense so far?
yes
!!!
How far is it from (4,0) to (2,0) ?
(2,0) ?? i dont know xD
|dw:1437263870646:dw|
do you see how they are 2 units apart ? |dw:1437263878771:dw|
yes i see that
I'll brb, I have to finish some of my tests
|dw:1437264011320:dw|
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it makes no sense to me
The general circle equation is this |dw:1437264186407:dw|
(h,k) is the center r is the radius
we're told that the center is (4,0) so (h,k) = (4,0) meaning that h = 4, k = 0 |dw:1437264245046:dw|
|dw:1437264266650:dw|
r = 2 was just found and shown above |dw:1437264297767:dw|
|dw:1437264320072:dw|
also notice how \(\Large (y-0)^2\) turned into \(\Large y^2\)
how did you get 2 squared, can you explain that?
that's what i don't get
well the right side of the general circle equation has \(\Large r^2\) I replaced r with 2 because the radius is r = 2
r = 2 \[\Large r^2 = 2^2\]
|dw:1437264515515:dw|
ohh i see okay
so x - 4squared + ysquared = 2squared (aka 4)
yes \[\Large (x-4)^2 + y^2 = 4\]
make sure you use parenthesis don't say x-4^2 say (x-4)^2
okay thanks a lot , i get it now
4^2 means "4 squared"
thank you lots
You could also have used the equation of the circle to figure out the radius algebraicly. We knew \[ (x-4)^2+y^2=r^2 \]And we knew \((x,y)=(2,0)\) lie in the circle, so \[ (2-4)^2+0^2=r^2 \\ (-2)^2 = r^2 \implies |r| = 2 \]But since we know \(r\) is positive for it to be a circle, we pick \(r=2\).
:o
you're welcome

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