Question
Draw a circle of radius 6cm. From a point 10cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
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Answer
Given that Construct a circle of radius 6cm, and let point P = 10cm form its centre, construct the pair of tangents to the circle. Find the length of tangents. We follow the following steps to construct the given 
Step of construction:
Step I: First of all we draw a circle of radius AB = 6cm.
Step II: Make a point P at a distance of OP = 10cm, and join OP.
Step III: Draw a right bisector of OP, intersecting OP at Q.
Step IV: Taking Q as centre and radius OQ = PQ, draw a circle to intersect the given circle at T and T’.
Step V: Joins PT and PT’ to obtain the require tangents. Thus, PT and P'T' are the required tangents. Find the length of tangents.
As we know that $OT \perp PT$ and $\triangle OPT$ is right triangle. Therefore, $OT =6 cm$ and $PO =10 cm \ln \triangle OPT PT ^2= OP ^2$ $-O T^2=10^2-6^2=100-36=64 PT =\sqrt{64}$
$=8$ Thus, the length of tangents $=8 cm$
Step of construction:Step I: First of all we draw a circle of radius AB = 6cm.
Step II: Make a point P at a distance of OP = 10cm, and join OP.
Step III: Draw a right bisector of OP, intersecting OP at Q.
Step IV: Taking Q as centre and radius OQ = PQ, draw a circle to intersect the given circle at T and T’.
Step V: Joins PT and PT’ to obtain the require tangents. Thus, PT and P'T' are the required tangents. Find the length of tangents.
As we know that $OT \perp PT$ and $\triangle OPT$ is right triangle. Therefore, $OT =6 cm$ and $PO =10 cm \ln \triangle OPT PT ^2= OP ^2$ $-O T^2=10^2-6^2=100-36=64 PT =\sqrt{64}$
$=8$ Thus, the length of tangents $=8 cm$
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