More Tangent Circles
by
Susan Sexton
For this assignment I will
construct the circles that are tangent to a given circle at a point on the
circle and a given line.
Construction of First Circle:
Given circle K whose center
is at K, point P that lies on circle K and line L, construct the line through K
and P that intersects line L at point A.
Construct the line
perpendicular to line KP through point P.
This line will intersect line L at point B.
Construct the circle that
whose center is point B and radius is BP.
Circle B will intersect line L at points C and D.
Construct the line
perpendicular to line L at point C.
This line will intersect line KP at point E.
Construct the circle whose
center is E and whose radius is EP.
Circle E is tangent to circle K at point P and tangent to line L at point C.
Discussion:
The construction can be
linked back to finding the tangent lines to a given circle from a point in the
exterior of the circle. (Discussed
in Assignment 6.) If we can find
the exterior point to the desired circle then we are half – way to
constructing the desired circle.
Point P lies on the tangent to both circles (circle K and the desired
circle) and intersects line L at point B so B is the exterior point. Therefore segment PB will be equal to the
distance from point B to the point of tangency between line L and the desired
circle. The point of tangency will
be perpendicular to the radius of the circle that lies on line KP.
Construction of the Second
Circle:
Construct the line
perpendicular to line L through K and intersects circle K at point G.
Construct the line through P
and G that intersects line L at point F.
Construct the line
perpendicular to line L through point F and intersects line KP at point H.
Construct the circle whose
center is point H and radius is HF. This circle will be tangent to circle K at point P and tangent to line L at point F.
Discussion:
Here the construction used similar triangles in finding the point of tangency (point F) along line L. The two similar triangles are ÆKPG and ÆHPF. Since KP = KG then the distance from the center of the desired circle (H) to point P should be equal to the distance of the perpendicular from H to line L. This perpendicular will intersect line L at the desired point of tangency. Line PG will also intersect line L at the desired point of tangency due to the similar triangles.
Here is a GSP sketch to watch
the circles in movement.