r2 = (2c)2 + (b-r)2
r2 = 4c2 + b2 -2br+r2
r = 4c2/2b + b/2
r = ((2a)2 -b2 )/2b + b/2
r = 2a2/ b
故在滑鼠任意點 (x,y) 與已知點 (fx,fy) 可求得底邊點 (r, fy)。則 r 點 x 座標值為 fx - 2a2/ b,其中:
a = 1/2 ( sqrt((fy-y)2 + (fx-x)2 )
b = fx-x
所以,就這樣
r2 = (2c)2 + (b-r)2
r2 = 4c2 + b2 -2br+r2
r = 4c2/2b + b/2
r = ((2a)2 -b2 )/2b + b/2
r = 2a2/ b
故在滑鼠任意點 (x,y) 與已知點 (fx,fy) 可求得底邊點 (r, fy)。則 r 點 x 座標值為 fx - 2a2/ b,其中:
a = 1/2 ( sqrt((fy-y)2 + (fx-x)2 )
b = fx-x
所以,就這樣