Problems on Common Tangents to Two Circles

Here we will solve different types of  problems on common tangents to two circles.

1. There are two circles touch each other externally. Radius of the first circle with centre O is 8 cm. Radius of the second circle with centre A is 4 cm. Find the length of their common tangent BC.

Common Tangent to Two Circles

Solution:

Join O to A and B. Join A to C. Draw DA ⊥ OB.

Now DA = BC, as they are opposite sides of rectangle ACBD.

OA = 8 cm + 4 cm

     = 12 cm.

OD = 8 cm – 4cm

     = 4 cm.

Therefore, DA = \(\sqrt{OA^{2} - OD^{2}}\)

                     = \(\sqrt{12^{2} - 4^{2}}\) cm

                     = \(\sqrt{144  - 16}\) cm

                     = \(\sqrt{128}\) cm

                     = 8√2 cm

Therefore, BC = 8√2 cm.


2. Prove that a transverse common tangent drawn to two circles divides the line joining their centres into the ratio of their radii.

Problems on Transverse Common Tangents

Solution:

Given: Two circles with centres O and P, and radii OX and PY respectively. The transverse common tangent XY touches them at X and Y respectively. XY cuts OP at T.

To prove: \(\frac{OT}{TP}\) = \(\frac{OX}{PY}\).

Proof:

Statement

Reason

1. In ∆XOT and ∆YPT,

(i) ∠OXT = ∠PYT = 90 °

(ii) ∠OTX = ∠PTY.

1.

(i) Tangent ⊥ Radius.

(ii) Vertically opposite angles.

2. ∆XOT ∼ ∆YPT

2. By A – A criterion of similarity.

3. Therefore, \(\frac{OT}{TP}\) = \(\frac{OX}{PY}\). (Proved)

3. Corresponding sides of similar triangles are proportional.






10th Grade Math

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