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.

Equal Transverse Common Tangents

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.

Length of a Transverse Common Tangent


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

Properties of Transverse Common Tangents

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.

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