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
Construction Procedure:
The given circle’s tangent can be constructed as follows.
1. Draw a circle with centre “O” having a radius of 3cm.
2. Construct a diameter of a circle that extends 7 cm from the circle’s centre and is marked as P and Q.
3. Now make a perpendicular bisector of the line PO and the midpoint is marked as M.
4. Draw a circle with centre M and MO as radius.
5. Join the points PA and PB where the circle with radius MO intersects the 3cm circle.
6. Now the required tangents are PA and PB.
7. In the similar way, we can draw the tangents, from the point Q.
8. From that, the required tangents are QC and QD.
Justification:
The given problem’s construction can be justified by proving that PQ and PR are tangents to a circle of radius 3 cm with centre O.
Join OA and OB to prove this.
From the above construction,
In the semi-circle ∠PAO is an angle.
As we know that the angle in a semi-circle is a right angle, we may write it as,
∠PAO = 90°
As a result
⇒ OA ⊥ PA
PA must be a tangent of the circle since OA is the radius of the circle with a radius of 3 cm. Similarly, we may now prove that PB, QC, and QD are the circle’s tangents.
As a result, the above construction justified.