Question

In the figure, the pair of tangents AP and AQ drawn from an external point A to a
circle with centre O are perpendicular to each other and length of each tangent is 5
cm. Find the radius of the circle.
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

Tangent at P and Q of the circle with centre O intersect at A.

Now,

OP is radius of circle and PA is tangent to a circle.

We know, Radius and tangent are perpendicular to each other.

So, OP ⊥ AP

⇛ ∠OPA = 90°

Also,

OQ is radius of circle and QA is tangent to a circle.

We know, Radius and tangent are perpendicular to each other.

So, OQ ⊥ AQ

⇛ ∠OQA = 90°

We know, Length of tangents drawn from external point toa circle are equal.

⇛ AP = AQ = 5 cm

Also,

Given that tangent AP and AQ intersecting at A such that ∠PAQ = 90°

Now, we have

∠OPA = 90°

∠OQA = 90°

∠PAQ = 90°

In quadrilateral OPAQ

We know, sum of all interior angles of a quadrilateral is 360°

So, ∠POQ + ∠OPA + ∠PAQ + ∠OQA = 360°

⇛ ∠POQ + 90° + 90° + 90° = 360°

⇛ ∠POQ = 90°

As each angle of OPAQ is 90°

It means OPAQ is a rectangle.

Now, we proved above AP = AQ

⇛ AP = AQ = OP = OQ = 5 cm

⇛ Radius of circle is 5 cm.

Alternative Method :-

As AP ⊥ AQ

⇛ ∠PAQ = 90°

We know, tangents are equally inclined to the line segment joining the centre and external point.

So, ∠PAO = ∠QAO = 45°

So, In ∆ OAP

Using Trigonometry,

$\rm :\longmapsto\:tan45 \degree \: = \: \dfrac{OP}{AP}$

$\rm :\longmapsto\:1 \: = \: \dfrac{OP}{5}$

$\bf\implies \:OP \: = \: 5 \: cm$

Hence, Radius of circle is 5 cm.

Additional Information :-

1. Radius and tangent are perpendicular to each.

2. One and only one tangent can be drawn from a point on circle.

3. Tangents are equally inclined to the line segment joining the center and external point.

4. Length of tangent drawn from external point are equal.

5. Angle in alternate segment are equal.