A circle passes through the point (12,0), and is tangent to the y-axis at the point (0,3), as shown. Find the radius of the circle.
Thanks!
~Noori
Let center ( r,3) radius:r
(x-r)^2+(y-3)^2=r^2 plug in (12,0)
(12-r)^2+(-3)^2=r^2 r=45/8
Ok apart from the final step. Should end up with r = 51/8
Ooh, thank you! That was very helpful. I thought the steps were right, so thank you!
A(0, 3) B(12, 0) origin O center of a circle C
1/ Line segment AB = sqrt(AO2 + BO2)
2/ Angle (ABO) = arctan(AO / BO) ∠ABO = ∠BAC
3/ Radius AC = (AB/2) / cos∠BAC
Sorry, I don't know trigonometry yet, is there a geometric solution?
+218 
