Question 165

A point Q is 13 crn from the centre of a circle. The length of the tangent drawn from Q to a circle is 12 cm. The distance of Q from the nearest point of the circle is

Solution

Here, O is the centre, QB is tangent = 12 cm

OQ = 13 cm

$$\angle$$OBQ = 90°

From, $$\triangle$$OBQ

OB = $$\sqrt{(OQ)^2-(BQ)^2}$$

= $$\sqrt{13^2-12^2}$$ = 5 cm

Now, OA = OB = 5 cm (radii)

=> Shortest distance = AQ = OQ - OA

= 13-5 = 8 cm

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