Question

two circles of radii 5cm and 3cm intersect at two points and the distance between their Centre is 4 cm find the length of the common chord

Answer 1

length of course=8cm

Answer 2

Answer:

Given :

Two intersecting circles with centres O and O'.

• OA = 5cm
• AO' = 3 cm
• OO' = 4 cm

To prove :

AB = 2AL ( OO' is the perpendicular bisector of AB )

Proof :

In ∆ AOL, using Pythagoras theorem

$\\ \sf \: {(5)}^{2} = {x}^{2} + {y}^{2} \\ \\ \\ \implies \sf \: {x}^{2} + {y}^{2} = 25 \qquad \: \: \: \: \quad( \: {eq}^{n} \: 1 \: )$

In ∆ ALO' , using Pythagoras theorem

$\\ \sf (3) {}^{2} = {y}^{2} + (4 - x) {}^{2} \\ \\ \\ \implies \sf \: 9 = {y}^{2} + 16 + {x}^{2} + 8x \\ \\ \\ \implies \sf \: 9 = {y}^{2} + {x}^{2} + 16 - 8x \\ \\ \\ \implies \sf \: 9 = 25 + 16 - 8x \\ \\ \\ \implies \sf \: 9 = 41 - 8x \\ \\ \\ \implies \sf \: - 8x = 9 - 41 \\ \\ \\ \implies \sf \: - 8x = - 32 \\ \\ \\ \implies \sf \red{x = 4}$

Put x = 4 in eqⁿ 1

$\\ \sf \: (4) {}^{2} + (y) {}^{2} = 25 \\ \\ \\ \implies \sf \: 16 + {y}^{2} = 25 \\ \\ \\ \implies \sf \: {y}^{2} = 25 - 16 \\ \\ \\ \implies \sf \: {y}^{2} = 9 \\ \\ \\ \implies \sf \pink{y = 3 \: \: cm}$

or, AB = 2AC = 2y = 2 × 3 = 6 cm