Can you show that this is a circle?

Edit : More clearly

Any 3D circle can be represented by an equation like r = r₀+ucosθ+vsinθ. u & v are orthogonal vectors in the plane of the circle both of length equal to the radius and r₀ is the centre. Conversely this equation represents a circle only if |u=|v| and u•v=0. The circle lies in the plane containing r₀ with normal uXv.

So if a parameterization is a circle it will be expressable in this form with |u|=|v| & u•v=0.

Set k² = (a²+b²+c²)/(2a²+(b−c)²), t=ktan(θ)

x = ( (b−c)ktanθ+atan²θ ) / sec²θ = ½a + ½(b−c)ksin(2θ) − ½acos(2θ)

y = ½b + ½(−a)ksin(2θ) – ½bcos(2θ)

z = ½c + ½(a)ksin(2θ) – ½ccos(2θ)

For this u = (k/2)( b−c, −a, a ) and v = ½( −a, −b, −c )

u•v = 0 and |u|² = ¼k²((b−c)²+2a²) = ¼(a²+b²+c²), |v|² = ¼(a²+b²+c²) so |u|=|v|

Centre is ½(a,b,c) and radius is ½√(a²+b²+c²) as already shown

The other curve

x = 2(t+20)(2t+19) / (50+12t+t^2),

y = (250+172t+19t^2) / (50+12t+t^2),

z = -2(t+20)(4t+17) / (50+12t+t^2)

Write denominator as (t+6)²+14 and set t+6=√14tanθ so denom = 14sec²θ

2(t+20)(2t+19) = 2(√14*tanθ+14)(2√14*tanθ+7) = 56tan²θ+70√14tanθ+196

∴ x = 2(1–cos(2θ)) + (5/2)√14sin(2θ) + 7(1+cos(2θ))

19t²+172t+250 =19(t+6)²−56(t+6)−98 = 19*14*tan²θ−56*√14tanθ−98

∴ y = (19/2)(1−cos(2θ)) − 2√14*sin(2θ) – (7/2)(1+cos(2θ))

−2(t+20)(4t+17) = −2(√14tanθ+14)(4√14tanθ−7) = −112tan²θ−98√14tanθ +196

∴ z = −4(1−cos(2θ)) – (7/2)√14sin(2θ) + 7(1+cos(2θ))

u = ( 5, −13, 11 ), v = √14*( 5/2, −2, −7/2 )

u•v = 0, |u| = |v| = √315 so curve is a circle and a = (9,6,3)

I don't know an easy way. But by brute force

a(b+c) x(t) + (bc-a^2-c^2) y(t) + (bc-a^2-b^2) z(t) = 0

So your curve is planar and passes through (0,0,0) when t=0.

Also, by brute force, (x(t)-a/2)^2 + (y(t)-b/2)^2 + (z(t)-c/2)^2 = (a^2+b^2+c^2)/4

so it is a circle, and I see now that you give us that in additional details, so my brute force efforts were for nothing.

Perhaps someone else can give the "right" way to do this problem.