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
PT^{2} = OP^{2} – OT^{2} (by Pythagoras’ Theorem)
⟹ PT^{2} = d^{2} – (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. |
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