Question

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

Answer 1

Step-by-step explanation:

Let the common chord be AB and P and Q be the centers of the two circles.

∴AP=5cm and AQ=3cm.

PQ=4cm ....given

Now, segPQ⊥chord AB

$\tt \: AR=RB= \frac{1}{2} AB$

from center to the chord, bisects the chord

Let PR=xcm, so RQ=(4−x)cm

In △ARP,

$\longrightarrow \tt \: {AP}^{2} = {AR}^{2} + {PR}^{2 } \\ \tt \: {AR}^{2} = {5}^{2} - {x}^{2} ----(i) \\ \tt \: In △ARQ, \\ \tt \: {AQ}^{2} = AR ^{2} + {QR}^{2} \\ \tt AR ^{2} = {(3)}^{2} - {(4 - x)}^{2} ---(ii) \\ \tt {5}^{2} - {x}^{ 2} = {3 - {(4 - x)}^{2} } ---- \: from \: (i \: and \: ii) \\ \tt 25 - {x}^{2} = 9 - (16 - 8x + {x)}^{2} \\ \tt \: 25 - {x}^{2} = - 7 + 8x - {x}^{2} \\ \tt 8x = 32 \\ \therefore \: \tt \: x = 4 \\ \tt Substitute \: in \: eq \: (1) \: we \: get, \\ \\ \tt \: AR ^{2} = 25 - 16 = 9 \: \\ \therefore \: \tt AR = 3 \\ \therefore \tt\:AB = 2 \times AR = \blue{6 \: cm}$

So, length of common chord AB is 6cm.