Problem & solution (1): Draw a circle which passes through a given point P and tangents on rays of given angle

Problem: Let xAy be a given angle and P be a given point inside of ∠xAy. Construct a circle which passes through point P and tangents on rays Ax, Ay.

Solution:

Suppose that a circle C1(O, OP) has been constructed (see a picture). We have following remarks:

1) Center O of drawn circle lies on a bisector Az of an angle xAy.

2) Let O1 be an arbitrary point on Az and C2(O1, O1J) be a circle of center O1 which tangents on Ay, then C2 also tangents on Ax.

3) Let I, J be tangent points of C1, C2 on Ay, then OI and O1J are perpendicular to Ay => OI // O1J => AO1 / AO = O1J / OI (a)

4) Draw a straight line which passes O1 and is parrallel to OP. This line intersects AP at point Q. We have O1Q / OP = AO1 / AP, on other side OP = OI => O1Q / OI = AO1 / AO (b)

5) From (a) and (b) we have O1Q = O1J => Q is on the circle C2

By the above, we can construct the circle C1 as bellowing steps:

1) Draw a bisector Az of an angle xAy, take an arbitrary point O1 on Az and draw a circle C2 of center O1 which tangents on Ay.

2) Let Q be a intersection point of C2 and straight line passing through A and P. From P draw a straight line which is parallel to O1Q. Let O be an intersection of this new line with Az.

A circle of center O, radius OP is tangent on Ax and Ay and passes through point P!

Proof: Draw OI ⊥ Ay, I ∈ Ay and O1J ⊥ Ay, J ∈ Ay => ∆ AOI ∾ ∆AO1J and ∆ AOP ∾ ∆AO1Q => OP / AO = O1Q / AO1 = O1J / AO1 = OJ / AO => OP = OJ = P ∈ C2. In other side C1 tangents on Ay => C1 tangents on Ax = C1 is a required circle!

Justification: the problem has two distinct solutions if P is not on Ax, Ay. In other cases we have one unique solution!

3.2.2014

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