Draw a circle of radius 4 cm.


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.


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

  1. Take a point 0 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.

Justificatign in $\triangle O A P$, we have

$O A=O P=4 \mathrm{~cm}$ ( $\because$ Radius)

Also, $A P=4 \mathrm{~cm}$ ( $\because$ Radius of circle with centre $A$ )

$\therefore \Delta O A P$ is equilateral

$\Rightarrow \quad \angle P A O=60^{\circ}$

$\Rightarrow \quad \angle B A P=120^{\circ}$

In $\triangle B A P$, we have

$B A=A P$ and $\angle B A P=120^{\circ}$

$\therefore \quad \angle A B P=\angle A P B=30^{\circ}$

$\Rightarrow \quad \angle P B Q=60^{\circ}$

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

$O A=4 \mathrm{~cm}$.

2. At $O$ construct radii $O A$ and $O B$ such that to $\angle A O B$ equal $120^{\circ}$ i.e., supplement of the

angle between the tangents.

3. Draw perpendiculars to $O A$ and $O B$ at $A$ and $B$, respectively. Suppose these

perpendiculars intersect at $P$. Then, $P A$ and $P B$ are required tangents.


In quadrilateral $O A P B$, we have

$\angle O A P=\angle O B P=90^{\circ}


$\angle A O B=120^{\circ}$

$\therefore \quad \angle O A P+\angle O B P+\angle A O B+\angle A P B=360^{\circ}$

$\Rightarrow \quad 90^{\circ}+90^{\circ}+120^{\circ}+\angle A P B=360^{\circ}$

$\therefore \quad \angle A P B=360^{\circ}-\left(90^{\circ}+90^{\circ}+120^{\circ}\right)$


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