Question

Draw a circle of radius 4 cm. Construct a pair of tangents to it, the angle between which is `60^(@)`. 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 centre draw a circle of radius AO=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 `DeltaOAP`, we have
OA=OP=4cm (`:.` Radius )
Also, AP=4cm (`:.` Radius of circle with centre A )
`:. DeltaOAP` is equilateral
`implies anglePAO=60^(@)`
`implies angleBAP=120^(@)`

In `DeltaBAP`, we have
`BA=AP " and " angleBAP=120^(@)`
`: angleABP= angleAPB =30^(@)`
`implies anglePBQ=60^(@)`
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=4 cm.
2. At O construct radii OA and OB such that to `angleAOB` equal `120^(@)` i.e., supplement of the angle between the tangents.
3. Draw perpendicular to OA and OB at A and B, respectively. Suppose these perpendicular intersect at P. Then, PA and PB are required tangents.

Justification
In quadrilateral OAPB, we have
`angleOAP=angleOBP=90^(@)`
and `angle AOB=120^(@)`
`:. angleOAP+angleOBP+angleAOB+angleAPB=360^(@)`
`implies 90^(@)+90^(@)+120^(@)+angleAPB=360^(@)`
`:. angle APB=360^(@)-(90^(@)+90^(@)+120^(@))`
`=360^(@)-300^(@)=60^(@)`