Question

Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is ${60}^{\circ }$. Also justify the construction. Measure the distance between the centre of the circle and the point of intersection of tangents.

Answer 1

In order to draw the pair of tangents, we follow the following steps

Steps of construction
1. Take a point O on the plane of the paper and draw a circle of radius OA = 4 cm
2. Produce OA to B such that OA = AB = 4 cm
3.Taking A as the center draw a circle of radius OA = AB = 4 cm
Suppose it cuts the circle drawn in step 1 at P and Q
4. Join BP and BQ to get desired tangents.

Justification in $\Delta OAP$, we have

OA = OP = 4 cm ($\because$ Radius)

Also, AP = 4 cm ($\because$ Radius of circle with centre)

$\Delta OAP$ is equilateral

$\Rightarrow \angle PAO={60}^{\circ }$

$\Rightarrow \angle BAP={120}^{\circ }$

In $\Delta BAP$ we have

BA = AP and $\angle BAP={120}^{\circ }$

$\therefore \angle ABP=\angle APB={30}^{\circ }$

$\Rightarrow \angle PBQ={60}^{\circ }$

Alternate method

Steps of construction
1. Take a point O on the plane of the paper and draw a circle with centre O and radius OA = 4cm
2. AT O construct radii OA and OB such that to $\angle AOB$ equal ${120}^{\circ }$ i.e supplement of the angle between the tangents.
3. Draw perpendicular to OA and OB at A and B respectively Suppose these Perpendicular intersects at P. Then PA and PB are required tangents.

Justification

In quadrilateral OAPB we have

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

And $\angle ABO={120}^{\circ }$

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

$\angle APB={360}^{\circ }-({90}^{\circ }+{90}^{\circ }+{120}^{\circ })$

$={360}^{\circ }-{300}^{\circ }={60}^{\circ }$