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anonymous
 one year ago
What is the general form of the equation of a circle with its center at (2, 1) and passing through (4, 1)?
x2 + y2 − 4x + 2y + 1 = 0
x2 + y2 + 4x − 2y + 1 = 0
x2 + y2 + 4x − 2y + 9 = 0
x2 − y2 + 2x + y + 1 = 0
anonymous
 one year ago
What is the general form of the equation of a circle with its center at (2, 1) and passing through (4, 1)? x2 + y2 − 4x + 2y + 1 = 0 x2 + y2 + 4x − 2y + 1 = 0 x2 + y2 + 4x − 2y + 9 = 0 x2 − y2 + 2x + y + 1 = 0

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DecentNabeel
 one year ago
Best ResponseYou've already chosen the best response.1standard form of circle: (x  h)^2 + (y  k)^2 = r^2 center is (h , k) and radius is r for your circle, the center (h , k) is (2 , 1) so it will look like: (x +2)^2 + (y  1)^2 = r^2 to find the radius squared, plug in the (x,y) point you know is on the circle  (4,1) (4 +2)^2 + (1  1)^2 = r^2 2^2 + 0 = 4 = r^2 thus, the equation is: (x +2)^2 + (y  1)^2 = 4

DecentNabeel
 one year ago
Best ResponseYou've already chosen the best response.1x^2+4+4x+y^2+12y=4 x^2+y^2+4x2y+1=0 so second opton is the answer

DecentNabeel
 one year ago
Best ResponseYou've already chosen the best response.1are you understand @king12233
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