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Note :

If center of the circle is not given, ( Ex 11.2, 7 )

 

We find its center first by

1.Taking any two non-parallel chords

2.And then finding the point of intersection of their perpendicular bisectors.

3.Point of intersection of perpendicular bisectors is the center of circle

 

Then you could proceed as Construction 11.3

  1. Chapter 11 Class 10 Constructions
  2. Serial order wise
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Transcript

Construction 11.3 To construct the tangents to a circle from a point outside it. We are given a circle with center O and point P outside the circle We need to draw tangents from point P to the circle Let’s follow these steps Steps of construction 1.Join PO. Make perpendicular bisector of PO Let M be the midpoint of PO. Note: To learn how to draw perpendicular bisector, check Construction 11.2 of Class 9 2.Taking M as centre and MO as radius, draw a circle. 3.Let it intersect the given circle at points Q and R. 4.Join PQ and PR. Then, PQ and PR are the required two tangents. Justification We need to prove that PQ and PR are the tangents to the circle. Join OQ and OR. Now, ∠PQO is an angle in the semi-circle of the blue circle And we know that, Angle in a semi-circle is a right angle. ∴ ∠PQO = 90° ⇒ OQ ⊥ PQ Since OQ is the radius of the circle, PQ has to be a tangent of the circle. (Since Tangent is perpendicular to radius) Similarly, PR is a tangent of the circle. Note : If center of the circle is not given, (Ex 11.2, 7) We find its center first by 1.Taking any two non-parallel chords 2.And then finding the point of intersection of their perpendicular bisectors. 3.Point of intersection of perpendicular bisectors is the center of circle Then you could proceed as Construction 11.3.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.