# Important Properties of Transverse Common Tangents

I. The two transverse common tangents drawn to two circles are equal in length.

Given:

WX and YZ are two transverse common tangents drawn to the two given circles with centres O and P. WX and YZ intersect at T.

To prove: WX = YZ.

Proof:

 Statement Reason 1. WT = YT. 1. The two tangents, drawn to a circle from an external point, are equal in length. 2. XT = ZT. 2. An in statement 1. 3. WT + XT = YT + ZT⟹ WX = YZ. (Proved) 3. Adding statements 1 and 2.

II. The length of a transverse common tangent to two circles is $$\sqrt{d^{2} – (r_{1} + r_{2})^{2}}$$, where d is the distance between the centres of the circles, and r$$_{1}$$ and r$$_{2}$$ are the radii of the given circles.

Proof:

Let two circles be given with centres O and P, and radii r$$_{1}$$ and r$$_{2}$$ respectively, where r$$_{1}$$ < r$$_{2}$$. Let the distance between the centres of the circles, OP = d.

Let WX be a transverse common tangent.

Therefore, OW = r$$_{1}$$ and PX = r$$_{2}$$.

Also, OW ⊥ WX and PX ⊥ WX, because a tangent is perpendicular to the radius drawn through the point of contact

Produce W to T such that WT = PX = r$$_{2}$$. Join T to P. In the quadrilateral WXPT, WT ∥ PX, as both are perpendiculars to WX; and WT = PX. Therefore, WXPT is a rectangle. Thus, WX = PT, as the opposite sides of a rectangle are equal.

OT = OW + WT = r$$_{1}$$  +  r$$_{2}$$.

In the right-angled triangle OPT, we have

PT2 = OP2 – OT2 (by Pythagoras’ Theorem)

⟹ PT2 = d2 – (r$$_{1}$$ + r$$_{1}$$)$$^{2}$$

⟹ PT = $$\sqrt{d^{2} – (r_{1} + r_{2})^{2}}$$

⟹ WX = $$\sqrt{d^{2} – (r_{1} + r_{2})^{2}}$$ (Since, PT = WX).

III. The transverse common tangents drawn to two circles intersect on the line drawn through the centres of the circles.

Given: Two circles with centres O and P, and their transverse common tangents WX and YZ, which intersects at T

To prove: T lies on the line joining O to P, i.e., O T and P lie on the same straight line.

Proof:

 Statement Reason 1. OT bisects ∠WTY⟹ ∠ATO = $$\frac{1}{2}$$∠WTY. 1. The tangents drawn to a circle from an external point are equally inclined to the line joining the point to the centre of the circle. 2. TP bisects ∠ZTX⟹ ∠XTP = $$\frac{1}{2}$$∠ZTX. 2.  As in statement 1. 3. ∠WTY = ∠ZTX. 3. Vertically opposite angles. 4. ∠WTO = ∠XTP. 4. From statement 1, 2 and 3. 5. OT and TP lie on the same straight line⟹ O, T, P are collinear. (Prove) 5. The two angles are forming a pair of vertically opposite angles.