Circle/Centre,P,Tanto,L
Circle with Explicit Centre Tangent to Line
Creates a circle with an explicit centre point P4, with radius computed so the circle is tangent to an existing line L2.
What It Does
The Circle/Centre,P,Tanto,L operator creates a circle with a specified centre point P4. The radius is computed automatically as the perpendicular distance from P4 to the line L2, making the circle tangent to L2.
Syntax
Circle/Centre, P4, Tanto, L2
Quick Input
P: 0
L: 0
Direction Selector
No direction selector required. The result is uniquely determined by the input geometry.
Worked Examples
Setup:
P0 = (50, 30) [explicit centre]
L0 = Horizontal line at Y=0
Circle/Centre, P0, Tanto, L0
→ C0: centre=(50,30)
radius = perpendicular dist from (50,30) to Y=0
= 30
C0 is tangent to L0 at (50, 0)
Setup:
P1 = (80, 60)
L1 = Vertical line at X=100
Circle/Centre, P1, Tanto, L1
→ C1: centre=(80,60)
radius = perpendicular dist from (80,60) to X=100
= |100-80| = 20
C1 is tangent to L1 at (100, 60)
Analytic Rule
Centre: (cx, cy) = P4 coordinates
Radius: r = perpendicular distance from P4 to L2
For line ax + by + c = 0:
r = |a*cx + b*cy + c| / sqrt(a^2 + b^2)
Tangent point: perpendicular projection of P4 onto L2
Result: Exact analytic circle.
FAQ
What if P4 is on the line L2?
If P4 lies exactly on L2, the perpendicular distance is 0 and the radius would be 0. XyzPilot returns an error for this degenerate case.
Can I use this with an angled line?
Yes. The perpendicular distance formula works for any line orientation — horizontal, vertical, or angled.