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Answer :

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]

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

\(=\sqrt{64}=8cm \)

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- Q.1 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.
- Q.2 Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
- Q.3 Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q
- Q.4 Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°
- Q.5 Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
- Q. 6 Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and \(\angle \) B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
- Q.7 Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.

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