Ex 11.2, 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.
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
We need to prove that PA, PB, QC, QD are the tangents to the circle.
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