Posted by **Anonymous** on Friday, August 31, 2007 at 5:40pm.

how would you find an equation for a circle that passes through the origin and has intercepts equal to 1 and 2 on the x and y axes

- calculus -
**Count Iblis**, Friday, August 31, 2007 at 7:36pm
The general equation for a circle is:

(x-a)^2 + (y-b)^2 = R^2

R is the radius of the circle and the point (a,b) is the center. To solve the problem you just insert the coordinates of the three points that are known to be on the circle in the equation and solve for a, b and R:

Origin: x = 0, y = 0:

a^2 + b^2 = R^2

x= 1, y= 0:

(1-a)^2 +b^2 = R^2

x= 0, y= 2:

a^2 + (2-b)^2 = R^2

Subtract from the last two equations the first equation to obtain two equations one containing only a and the other containing only b.

- calculus -
**Anonymous**, Saturday, September 1, 2007 at 5:02pm
ok so i found:

(y-1)^2 + (x-.5)^2 - r^2

so how do you find r??

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