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Ex 11.2, 3
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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.

The tangent can be constructed on the given circle as follows.

- Taking any point O on the given plane as centre, draw a circle of 3 cm radius.
- Extend diameter of circle to a distance of 7 cm from centre on both sides.
- Let these points be P and Q. Where OP = OQ = 7 cm
- Bisect OR and OS. Let M and N be the mid-points of OP and OQ respectively.
- .Taking M as centre and MO as radius, draw a circle. Do the same for point N. Let it intersect at points A, B and C, D respectively
- Join PA, PB, PC and PD

Thus, PA, PB, QC, QD are the required tangents

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Justification:
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We need to prove that PA, PB, QC, QD
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are
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the tangents to the circle.
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Join OA and OB.

∠PAO is an angle in the semi-circle of the blue circle

And we know that angle in a

semi-circle is a right angle.

∴ ∠PAO = 90°

⇒ OA ⊥ PA

Since OA is the radius of the circle,

PA has to be a tangent of the circle.

Similarly, PB is a tangent of the circle.

Similarly, we can prove QC and QD are tangents