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# Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Steps to construct the required figure:
Step 1: Draw a circle with center O and radius =6cm and draw a point P such that OP=10cm.

Step 2: Draw a perpendicular bisector of OP and mark a point M such that M is the mid- point of OP.

Step 3: With center as M and radius PM=MO, draw a circle which cuts the given circle at S and T.

Step 4: Now join PS and PT.

Thus, we have the required tangents as PS and PT and also the lengths of each tangent is 8cm.

JUSTIFICATION:

Join OS

Now in the triangle PSO,we have

$$∠PSO=90°$$

$$∴ PS= \sqrt{{OP}^2 -{OS}^2}$$
[by pythagoras’ theorem]

$$=\sqrt{{(10)}^2-{(6)}^2}=\sqrt{100-36}$$

$$=\sqrt{64}=8cm$$