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

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➧ Let the radius of the two circles be 5 cm and 3 cm respectively whose centre’s are O and O'.

➧ Hence OA = OB = 5 cm

➾ O'A = O'B

➾ 3 cm

➧ OO' is the perpendicular bisector of chord AB.

➧ Therefore, AC = BC

➧ Given, OO' = 4 cm

➧ Let OC = x

➧ Hence O'C = 4 − x

➧ In right angled ΔOAC, by Pythagoras theorem OA²

➾ OC² + AC²

➾ 5² = x² + AC²

➾ AC² = 25 − x²à

➧ (1) In right angled ΔO'AC, by Pythagoras theorem O'A²

➾ AC² + O'C²

➾ 3² = AC² + (4 – x)²

➾ 9 = AC² + 16 + x² − 8x

➾ AC² = 8x − x² − 7 à

➧ (2) From (1) and (2),

➧ we get 25 − x² = 8x − x² − 7 8x = 32

➧ Therefore, x = 4

➧ Hence, the common chord will pass through the centre of the smaller circle, O' and hence, it will be the diameter of the smaller circle.

➧ AC² = 25 − x²

➾ 25 − 4²

➾ 25 − 16 = 9

➧ Therefore, AC = 3 m Length of the common chord,

➾ AB = 2AC

➾ 6 cm ...✔

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Answer 2

Solution:

Given parameters are:

• OP = 5cm
• OS = 4cm and
• PS = 3cm
• Also, PQ = 2PR

Now, suppose RS = x.

Consider the ΔPOR,

OP2 = OR²+PR²

⇒ 52 = (4-x)²+PR²

⇒ 25 = 16+x²-8x+PR²

∴ PR² = 9-x²+8x ...... (i)

Now consider ΔPRS,

PS² = PR²+RS²

⇒ 32 = PR²+x²

∴ PR² = 9-x² ............ (ii)

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By equating equation (i) and equation (ii) we get,

9 -x²+8x = 9-x²

⇒ 8x = 0

⇒ x = 0

Now, put the value of x in equation (i)

PR2 = 9-0²

⇒ PR = 3cm

∴ The length of the cord i.e. PQ = 2PR

So, PQ = 2×3 = 6cm

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