Question

24. Draw two concentric circles of radii 3 cm and 5 cm. Taking a point on outer circle
construct the pair of tangents to the other. Measure the length of a tangent and verify it
by actual calculation.​

Answer 1

Required Answer:-

Firstly make a rough sketch of the construction as said above with a pencil and no specific measurement. Now as you have a proper view of how's it going to look, let's start the construction..

Steps of construction:-

• Draw a circle of radius 3 cm. Now with the same centre draw another circle of radius 5 cm.
• Extend O to any point on the bigger circle (Let it be point P).
• Bisect OP perpendicularly. And find the midpoint. Name this point as O'.
• Now taking O' as the centre, draw a circle of radius O'O.
• Mark the points in which this circle intersects the smaller circle. Let these points be A and B.
• Join P to A and P to B.

$\therefore$ AP and BP are the required tangents.

Verification:-

Take a ruler and measure the length of PA/PB because tangents from same exterior point to the circle are equal in measure. It is 4 cm.

Now join O to A. ∆PAO is a right angled triangle, right angled at A. We have OA = 3 cm because it is the radius of the smaller circle and OP = 5 cm, it is the radius of bigger circle.

Using Pythagoras theoram,

OA² + AP² = OP²

AP = √(OP² - OA²)

AP = √(5² - 3²) cm

AP = √4² cm

AP = 4 cm

Hence, verified!!

Answer 2

Steps of construction :

1. Draw a circle with radius 3 cm and centre O. .

2. Draw another circle with radius 5 cm and same centre O.

3. Take a point P on the circumference of larger circle and join O to p.

4. Taking OP as diameter draw another circle which intersects the smallest circle at A and B.

5. Join A to P and B to P. Hence AP and BP are the required tangents.

Verification $\checkmark$

Using Pythagoras theoram,

OA² + AP² = OP²

AP = $\sf \sqrt{(OP² - OA²)}$

AP = $\sf \sqrt{(5² - 3²) cm}$

AP = $\sf \sqrt{4²\: cm}$

AP = 4 cm

${\boxed{\sf{Hence, verified}}}$