SOLUTION: Find the equation of the circle centered at (4,6) and passing through the point (2,2). Thank you I'm having a hard time with circle equations!

Algebra ->  Coordinate-system -> SOLUTION: Find the equation of the circle centered at (4,6) and passing through the point (2,2). Thank you I'm having a hard time with circle equations!      Log On


   

Question 530469: Find the equation of the circle centered at (4,6) and passing through the point (2,2).
Thank you I'm having a hard time with circle equations!
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 Start with the general equation of a circle.


%28x-4%29%5E2%2B%28y-6%29%5E2=r%5E2 Plug in h=4 and k=6 (since the center is the point (h,k) ).


%282-4%29%5E2%2B%282-6%29%5E2=r%5E2 Plug in x=2 and y=2 (this is the point that lies on the circle, which is in the form (x,y) ).


%28-2%29%5E2%2B%28-4%29%5E2=r%5E2 Combine like terms.


4%2B16=r%5E2 Square each term.


20=r%5E2 Add.


So because h=4, k=6, and r%5E2=20, this means that the equation of the circle with center (4,6) that goes through the point (2,2) is


%28x-4%29%5E2%2B%28y-6%29%5E2=20.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find the equation of the circle centered at (4,6)
and passing through the point (2,2).
----
radius = sqrt[(6-2)^2+(4-2)^2] = sqrt[20)
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Form: (x-h)^2 + (y-k)^2 = r^2
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Your Problem:
(x-4)^2 + (y-6)^2 = 20
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Cheers,
Stan H.
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