![Ex 11.2, 3 - Chapter 11 Class 10 Constructions - Part 2](https://d1avenlh0i1xmr.cloudfront.net/82800212-074c-4e2a-9b9d-ab8443b054fb/slide15.jpg)
![Ex 11.2, 3 - Chapter 11 Class 10 Constructions - Part 3](https://d1avenlh0i1xmr.cloudfront.net/76aca433-f64e-4e87-9f1f-9e64db5130c4/slide16.jpg)
![Ex 11.2, 3 - Chapter 11 Class 10 Constructions - Part 4](https://d1avenlh0i1xmr.cloudfront.net/31fec844-9402-4c91-b8e9-4ff5ed006c16/slide17.jpg)
![Ex 11.2, 3 - Chapter 11 Class 10 Constructions - Part 5](https://d1avenlh0i1xmr.cloudfront.net/36ed5bab-a136-4623-83fc-cd68891cd50c/slide18.jpg)
Constructing tangents to a circle
Constructing tangents to a circle
Last updated at April 16, 2024 by Teachoo
Question 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. Steps of construction Draw a circle of radius 3 cm Draw diameter of circle, and extend it and mark points P and Q, 7 cm from the center Let’s first draw tangent from point P 3. Make perpendicular bisector of PO Let M be the midpoint of PO. 4 Taking M as centre and MO as radius, draw a circle. 5. Let it intersect the given circle at points A and B. 6. Join PA and PB. Now, we draw tangent from point Q Similarly we draw tangent from point Q ∴ QC and QD are the tangents from point Q Justification We need to prove that PA, PB, QC, QD are the tangents to the circle. Join OA, OB, OC and OD Now, ∠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, we can prove PB, QC, QD are tangents of the circle.