Test: Perpendicular From Centre To Chord - Class 9 MCQ

Consider AB as the diameter with centre O which bisects the chord CD at E

It is given that CE = ED = 8cm and EB = 4cm

Join the diagonal OC

Consider OC = OB = r cm

It can be written as OE = (r – 4) cm

Consider △ OEC

By using the Pythagoras theorem

OC² = OE² + EC²

By substituting the values we get

r² = (r – 4)² + 8²

On further calculation r² = r² – 8r + 16 + 64

So we get r² = r² – 8r + 80

It can be written as r² – r² + 8r = 80

We get 8r = 80 By division r = 10 cm

Let the radius of the circle centered at O and O' be 5 cm and 3 cm respectively.

OA = OB = 5 cm

O'A = O'B = 3 cm

OO' will be the perpendicular bisector of chord AB.

∴ AC = CB

⇒ 5² = AC² + x²

⇒ 25 − x² = AC² ...(1)

In ΔO'AC,

O'A2 = AC2 + O'C2

⇒ 3² = AC² + (4 − x)²

⇒ 9 = AC² + 16 + x² − 8x

⇒ AC² = − x² − 7 + 8x ... (2)

From equations (1) and (2), we obtain

25 − x² = − x² − 7 + 8x

8x = 32

x = 4

Therefore, the common chord will pass through the centre of the smaller circle i.e., O' and hence, it will be the diameter of the smaller circle.

AC² = 25 − x² = 25 − 42 = 25 − 16 = 9

∴ AC = 3 m

Length of the common chord AB = 2 AC = (2 × 3) m = 6 m

Using Pythagoras theorem:

OA² = OM² + AM²

AM² = OA² - OM²

AM² = 13² - 12²

AM² = 169 - 144

AM² = 25

AM = 5 cm

AB = 2 AM

AB = 2 x 5 cm = 10 cm

Consider a circle having center O with a chord. Let OA be the radius of the circle and AB be the chord. As given in the question, the radius of the circle is 5 cm and length of chord is 8 cm. Let the distance between the center of the circle and chord be OP. So, this can be shown diagrammatically as:

It is clear from the diagram that OP is perpendicular to AB. As OP is perpendicular to AB and passes through the center O, it will bisect the chord AB at P. Now the length of AP will be,
AP =1/2 × AB

AP = 1/2 × 8

AP = 4 cm
Since, triangle OPA is a right-angle triangle, we can easily apply the Pythagoras theorem which can be stated as b²+p² = h² where b, p and h are base, perpendicular and hypotenuse of the respective triangle.
In ΔOPA,

AP²+OP²=AO²

OP² = AO² − AP²

OP²= 5²−4²

OP = 3 cm
Therefore, the distance of the chord AB from the center is 3 cm.

In the right triangle OAP,

OA² = OP² +AP² (By Pythagoras theorem)

OA² = 4² + 3² (perpendicular from the centre of the circle bisects the chord , AP=3cm)

OA² = 25

OA = 5

Hence the radius of the circle is 5 cm.

Here OPA is right angle triangle

given OP = 16 cm

AO = 34 cm

AP = ?

using Pythagoras theorem :

AO² = OP² + AP²

34² = 16² + AP²

1156 = 256 + AP²

1156 - 256 = AP²

900 = AP² 

√900 = AP

30 = AP

Since AD is a straight line and O is at center of both circles

AD = AP + PD

AD = AP + AP

AD = 2AP

AD = 2 x 30

AD = 60 CM

Given length of common chord AB =12 cm

Let the radius of the circle with centre O is OA = 10 cm

Radius of circle with centre P is AP = 8 cm

From the figure, OP⊥AB

⇒AC = CB

∴AC = 6 cm   (Since AB=12 cm)

In ΔACP, 

AP² = PC²+AC²    [By Pythagoras theorem]

⇒ 8² = PC²+6²  

⇒ PC² = 64–36 = 28

 PC = 2√7​ cm

Consider ΔACO,

AO² = OC²+AC²   [By Pythagoras theorem]

⇒102 = OC2+62  

⇒OC2 = 100−36 = 64

⇒OC = 8 cm
From the figure, OP = OC+PC = 8+2√7​ cm.
Hence, the distance between the centres is (8+2√7​) cm.