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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