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Angle Between Two Tangents

Circles I - Concepts

Radius Chord Secant Tangent Difference

Relationship Between Radius of Circle And Tangent

Number of Tangents From a Point on a Circle

Length of Tangents to a Circle

Angle Between Two Tangents

Circle and Related Terms

Arc of a Circle

Chord and Segment of a Circle

Class - 10th IMO Subjects

Concept Explanation

Angle Between Two Tangents

Angle Between Two Tangents : Form an external point we can draw two tangents to a circle. In the figure we have an external point P we can draw two tangents PA and PB . $angle$ APB is the angle between two tangents,

ILLUSTRATION: Prove that the angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the centre.

Solution In fig., PA and PB are two tangents to the circle from an external point P.

Join OA and OB. We know that the tangent makes an angle of $90^0$ with the radius

$large angle OAP=angle OBP=90^{circ}$

Also, sum of the angles of a quadrilateral is $large 360^{circ}$ . Therefore,

$large angle AOB+angle OAP+angle APB+angle OBP=360^{circ}$

$large Rightarrow ;;;angle AOB+angle APB+90^{circ}+90^{circ}=360^{circ}$

$large Rightarrow ;;;;;;angle AOB+angle APB=180^{circ}$

Thus, $large angle APB$ is supplementary to the angle $large angle AOB$ .

ILLUSTRATION: In the figure prove that PQR is an equilateral triangle where O and P are centre points of the circles

SOLUTION: PA and PB are tangents from point P to the circle with centre O

$Rightarrow ;;;angle OAP= 90^0 ;;; and;;; angle OBP = 90^0$ [tangent makes a right angle with radius]

Also, OAPB is a quadrilateral and sum of the angles of a quadrilateral is $large 360^{circ}$ . Therefore,

$large angle AOB+angle OAP+angle APB+angle OBP=360^{circ}$

$large Rightarrow ;;;angle AOB+angle APB+90^{circ}+90^{circ}=360^{circ}$

$large Rightarrow ;;;;;;angle AOB+angle APB=180^{circ}$

As $large angle AOB = 120^0$ [Given ]

$large Rightarrow ;;;;;;120^0+angle APB=180^{circ}$

$large Rightarrow ;;;;;;angle APB=60^{circ}$

Now $large Delta$ PQR is an isosceles triangle as PQ = PR [ radii of the circle with centre P]

$large angle PQR = angle PRQ= x (suppose)$ [Angles opposite to equal sides are equal]

Also $large angle PQR + angle PRQ+ angle QPR= 180^0$ [Angle sum property of triangle PQR]

$large x + x+ 60^0= 180^0 ;;Rightarrow ;; 2x= 180^0-60^0= 120^0$

$large Rightarrow ;; x= 60^0$

As all the angles of $large Delta$ PQR are $large 60^0$ , Therefore $large Delta$ PQR is an equilateral triangle.

Sample Questions

(More Questions for each concept available in Login)

Question : 1

If tangents PA and PB from a point P to a circle with centre O are inclined to each other an angle of 70° , then find ∠POA.
A. 50	B. 60	C. 55	D. 70
Right Option : C
View Explanation

Question : 2

In the given figure, PA and PB are tangents from P to a circle with centre O. If ∠AOB = 130°, then find ∠APB
A. 45	B. 55	C. 50	D. 60
Right Option : C
View Explanation