Note :

If centre 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.

Then you could proceed as Construction 11.3 .

-v-

Construction 11.3:

To construct the tangents to a circle from a point outside it.

We are given a circle with centre O and a point P outside it. We have to construct the two tangents from P to the circle.

Steps of Construction :

1. Join PO and bisect it. Let M be the midpoint of PO.
2. Taking M as centre and MO as radius, draw a circle.
3. Let it  intersect the given circle at the 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.

∠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.

Similarly, PR is a tangent of the circle.

1. Class 10
2. Important Questions for Exam - Class 10

Transcript

To construct the tangents to a circle from a point outside it. We are given a circle with centre O and a point P outside it. We have to construct the two tangents from P to the circle. Steps of Construction: Join PO and bisect it. Let M be the midpoint of PO. Taking M as centre and MO as radius, draw a circle. Let it  intersect the given circle at the points Q and R. 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. ∠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. Similarly, PR is a tangent of the circle.

Class 10
Important Questions for Exam - Class 10