Question

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

Answer 1

Hint: Here to proceed with the solution we need to have a basic understanding on the concepts of tangents, tangent of circle, supplementary angles and sum of quadrilateral angles concept.

Complete step-by-step answer:

Here if we observe the diagram O is the centre of the circle and p is the external point.
Since we know that if a straight line that touches the circle at only one point then it is known as tangent.
Now by using the above concept, from the diagram we know that PA and PB are tangent to the circle.
As we know the radius of the circle is perpendicular to tangent.
Then here
$OA \bot PA$
Similarly we can say that
$OB \bot PB$
Now from this we can say that
$\angle OBP = {90^ \circ }$
$\angle OAP = {90^ \circ }$
From the diagram, we can represent a quadrilateral OAPB.
We also that sum of interior angles = ${360^ \circ }$
So from this we can say that
In a quadrilateral OAPB, sum of interior angles = ${360^ \circ }$
Where we can say that
$ \Rightarrow \angle OAP + OBP + \angle BOA + \angle APB = {360^ \circ }$
$ \Rightarrow {90^ \circ } + {90^ \circ } + \angle BOA + \angle APB = {360^ \circ }$ (Since $\angle OBP = {90^ \circ }$ and $\angle OAP = {90^ \circ }$ )
$
   \Rightarrow \angle BOP + \angle APB = {360^ \circ } - {90^ \circ } - {90^ \circ } \\
   \Rightarrow \angle BOP + \angle APB = {180^ \circ } \\
 $
Since we know that Supplementary angles are two angles with the sum of ${180^ \circ }$.
Hence it is proved that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment.

Note: In this problem by using the concept of tangents we know that PA and PB are tangent to the circle. Since the radius of circle is perpendicular to the tangent of circle then we found that $\angle OBP = {90^ \circ }$,$\angle OAP = {90^ \circ }$.Later by using the concept of sum of interior angles of a quadrilateral. We got supplementary angle where $\angle BOP + \angle APB = {180^ \circ }$.From using all these concepts we have proved that angle between the two tangent drawn from an external point to a circle is supplementary to the angle subtended by the line segment.