Question

AB is a chord of a circle with centre O. At B, a tangent PB is drawn such that its length is 24cm. The distance of P from centre is 26cm. If the chord AB is 16cm, find its distance from the centre.

Answer 1

Heya,

According to the question,
PB = 24cm
PO = 26cm
AB = 16cm

As, we know that when the centre is joined with the tangent is forms 90° with the tangent.
So,
We, will join OB
And,
Angle OBP = 90°

By applying Pythagoras theorem, we get,

OB² + BP² = PO²

OB² = PO² - BP²

OB² = (26)² - (24)²

OB = √676 - 576

OB = √100

OB = 10

So, radius i.e. the distance from the centre = 10cm.

Now, as we now know the radius then we can now find the distance between the cord and the centre of the circle.

Let us make a line OD from centre O to the cord.

Now, by applying Pythagoras theorem, we can find OD

DB = 16/2 = 8cm
OB = 10cm

So,
OD = √(OB)² - (DB)²
= √(10)² - (8)²
= √100 - 64
= √36
= 6cm

So, the distance of cord from the centre = 6 cm

Hope this helps....:)

Answer 2

Given :-

AB is a chord of circle with centre O  and PB = 24 cm, OP = 26 cm.

By Pythagoras theorem,

$\bf\huge OB = \sqrt{(26)^2 - (24)^2}$

$\sqrt{676 - 576}$

$\sqrt{100}$

= 10 cm

Now in ΔOBC

$\bf\huge BC = \frac{16}{2} = 8 cm$

OB = 10 cm

OC^2 = OB^2 - BC^2  

= 102 – 82

OC^2 = 36

OC = 6 cm

Distance of the chord from the centre  = 6 cm