How do I find Cx
CD = sq. root{ [(sq. root (16 - (Cx-1)^2) + 2] - Cy }^2
Cx is supposed to equal + or - [sq. root (16- (CD+1)^2 ] +1
point D is on the circle. the coordinates for point D are
$\displaystyle (x, \sqrt{16-(x-1)^2} + 2)$
point C has coordinates $\displaystyle (x,3)$
CD is the vertical distance between D and C ... the difference between the y-values.
$\displaystyle CD = [\sqrt{16-(x-1)^2} + 2] - 3$
$\displaystyle CD = \sqrt{16-(x-1)^2} - 1$
$\displaystyle CD + 1 = \sqrt{16 - (x-1)^2}$
$\displaystyle (CD + 1)^2 = 16 - (x-1)^2$
$\displaystyle (x-1)^2 = 16 - (CD + 1)^2$
$\displaystyle x-1 = \pm \sqrt{16 - (CD + 1)^2}$
$\displaystyle x = 1 \pm \sqrt{16 - (CD + 1)^2}$