We will solve some Problems on two tangents to a circle from an external point.
1. If OX any OY are radii and PX and PY are tangents to the circle, assign a special name to the quadrilateral OXPY and justify your answer.
Solution:
OX = OY, are radii of a circle are equal.
PX = PY, as tangents to a circle from an external point are equal.
Therefore, OXPY is a kite.
2. ∆XYZ is right angled at Y. A circle with centre O has been inscribed in the triangle. If XY = 15 cm and YZ = 8 cm, find the radius of the circle.
Solution:
Using Pythagoras’ theorem, we get
XZ = \(\sqrt{XY^{2} + YZ^{2}}\) = \(\sqrt{225 + 64 }\) cm = \(\sqrt{289}\) cm = 17 cm.
We draw OP ⊥ XY, OQ ⊥ YZ and OR ⊥ XZ.
Therefore, OP = OQ = OR = r, where r is the radius of the circle.
PYQO is a square.
Therefore, PY = YQ = r.
Therefore, XP = 15 cm – r and QZ = 8 cm – r.
Now, tangents drawn to a circle from an external point are equal.
Therefore, XR = XP = 15 cm – r and RZ = QZ = 8 cm – r.
But XR + RZ = XZ
⟹ 15 cm – r + 8 cm – r = 17 cm
⟹ 23 cm – 2r = 17 cm
⟹ 2r = 23 cm – 17 cm
⟹ 2r = 6 cm
⟹ r = 3 cm.
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