Question

Draw a circle of radius 6cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths. Verify by using Pythagoras Theorem.

Answer 1

HELLO DEAR,

[figure is in the attachment]

A pair of tangents to the given circle can be constructed as follows.

Step 1

Taking any point O of the given plane as centre, draw a circle of 6 cm radius. Locate a point P, 10 cm away from O. Join OP.

Step 2

Bisect OP. Let M be the mid-point of PO.

Step 3

Taking M as centre and MO as radius, draw a circle.

Step 4

Let this circle intersect the previous circle at point Q and R.

Step 5

Join PQ and PR. PQ and PR are the required tangents.

The lengths of tangents PQ and PR are 8 cm each.

Justification

The construction can be justified by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 6 cm). For this, join OQ and OR.

∠PQO is an angle in the semi-circle. We know that angle in a semi-circle is a right angle.

∴ ∠PQO = 90°

⇒ OQ ⊥ PQ

Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR is a tangent of the circle.

now,

justification by Pythagoras theorem;

IN ∆ OPQ,

OP²  = PQ² + OQ²

(10)² = PQ² + (6)²

100 - 36 = PQ²

PQ = √64

PQ = 8

I HOPE ITS HELP YOU DEAR,

THANKS

Answer 2

Step1: Draw circle of radius 6cm with center A, mark point C at 10 cm from center

Step 2: find perpendicular bisector of AC

Step3: Take this point as center and draw a circle through A and C

Step4:Mark the point where this circle intersects our circle and draw tangents through C.

Length of tangents = 8cm
AE is perpendicular to CE (tangent and radius relation)
In ΔACE
AC becomes hypotenuse
AC² = CE²  + AE²
10² = CE² + 6²
CE² = 100-36
CE²  = 64
CE = 8cm