Question

Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.​

Answer 1

Answer:

2cm because 5cm and3cm intersect at two points

Answer 2

CONCEPT :

★ Let the common chord AB in two circles having P and Q as the center. AP is the radius of the first circle having length 5cm and AQ is the radius of the other circle having length 3cm. Now, segment PQ is perpendicular to the chord AB using the theorem “The perpendicular from center to the chord, bisect the chord”. Therefore, the length of AR is the half of the length AB using the theorem. After that using Pythagoras Theorem in the both triangle ARP and triangle ARQ. We get the value of length of AR, so the length of AB is half of the length of AR.

SOLUTION :

: $\implies$ Let the common chord be AB. P and Q are the centers of circles.

:$\implies$ The length of AP is given in the question, AP = 5 cm

: $\implies$ The length of PQ is also given in the question, PQ =4 cm

: $\implies$ Using the theorem “The perpendicular from center to the chord, bisect the chord”.

Hence, the segment PQ is perpendicular to the chord AB.

AR = RB = 1 / 2 AB

: $\implies$ Let the length of PR be x cm.

So, the length of RQ = ( 4 - x ) cm

: $\implies$ Now, consider the triangle ARP and apply Pythagoras Theorem in it:

In triangle ARP,

: $\implies$ AP² = AR² + PR²

: $\implies$ AR² = (5) ² - x²

: $\implies$ AR² = 25 - x² ... (i)

Applying the Pythagoras Theorem in the triangle ARQ,

: $\implies$ AQ² = AR² + QR²

: $\implies$ AR² = (3) ² - (4 -x ) .. (ii)

using equation 1 and 2 we get ,

∴ 5² - x² = 3² - ( 4 - x ) ²

25 - x ² = 9 - ( 16 - 8 x + x² )

We extend the ( 4 - x ) ² using the identities

( a - b )² = a ² + b² - 2 AB

solving further

25 - x ² = - 7 + 8 x - x ²

32 = 8x

∴ 32 = 8x

∴ x = 4

Putting the value of x, in equation number 1, we get:

AR ² = 25 -16

∴ AR = 3 cm

∴ AB = 2 x AR = 2 x 3

∴ AR = 6 CM

• Hence, the length of the common chord AB is 6cm.