Tangent to a Circle

Lesson Progress
0% Complete

All the tangents of a circle are perpendicular to the radius through the point of contact of that tangent.

OP is the radius of the circle and Q is any point on the line XY which is the tangent to the circle. As OP is the shortest line of all the distances of the point O to the points on XY. So OP is perpendicular to XY. Hence, OP \bot XY


Find the radius of the circle, if the length of the tangent from point A which is 5 cm away from center is 4 cm.


As we know that the radius is perpendicular to the radius, so the \triangle ABO is a right angle triangle.

Given, AO = 5 cm and AB = 4 cm We can use Pythagoras theorem here

OA2 = OB2 + AB2

OB= OA2 – AB2

= 52 – 42

= 25 – 16

OB= 9

OB = 3 So the radius of the given circle is 3 cm


If two tangents PA and PB are drawn to a circle from a point P with centre O and OP is equal to the diameter of the circle then show that triangle APB is an equilateral triangle.


Given, AP is a tangent to the circle.

Therefore, OA \bot AP (Tangent is perpendicular to the radius through the point of contact)

\angle OAP = 90^o

In \triangle OAP,

\sin\angle OPA=\dfrac{OA}{OP}=\dfrac{r}{2r}\ (OP = \text{diameter} )

So \sin \angle OPA=\dfrac{1}{2}=30^o

Also \angle =30^o

Now, \angle APB=\angle OPA+\angle OPB=30^o+30^o=60^o

In \triangle PAB,PA=PB (length of the tangents from the external point is equal)

\angle PAB=\angle PBA (Angles opposite to equal sides are equal)

\angle PAB+\angle PBA+\angle APB=180^o (Due to angle sum property)

\angle PAB+\angle PBA=180^o-60^o

2\angle PAB=120

\angle PAB=60^o

As \angle PAB=\angle PBA=\angle APB=60^o

Hence, PAB is an equilateral triangle.

Scroll to Top