Circle

Circle touch two parallel lines p and q; and its center lies on a line a, which is secant of lines p and q.

Write the equation of circle and determine the coordinates of the center and radius.

p: x-10 = 0
q: -x-19 = 0
a: 9x-4y+5 = 0


Result

xS =  -4.5
yS =  -8.88
r =  NAN

Solution:

A=pa=[10;0] B=qa=[19;0] S=AB2=[4.5;8.88] tp;tq;St t:+y+1=0 T=pt=[10;NAN] A = p \cap a = [10; 0] \ \\ B = q \cap a = [-19; 0] \ \\ S = \dfrac{ AB }{2} = [-4.5; -8.88] \ \\ t \perp p ; t \perp q; S \in t \ \\ t: +y +1 = 0 \ \\ T = p \cap t = [10; NAN] \ \\
yS=(23.75+(41.5))/2=718=8.88y_{ S }=(23.75+(-41.5))/2 = - \dfrac{ 71 }{ 8 } = -8.88
r=SA=(xSxT)2+(ySyT)2=NAN  (x+4.5)2+(x+8.88)2=NAN2 r = |SA| = \sqrt{ (x_S-x_T)^2+(y_S-y_T)^2 } = NAN \ \\ \ \\ (x +4.5)^2+(x +8.88)^2 = NAN^2 \ \\



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