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.
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.
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. |
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