Two circles of radii 10 cm and 8 cm intersect at two points and the length of the common chord is 12 cm. Find the distance between their centres.
Proof: In circle 'a' OP is radius. PQ is chord, OX⊥PQ ⇒PX=12PQ→⊥ from centre to chord bisects chord PQ=12 cm ⇒PX=6cm PO−PX=12OX ⇒10−6=4cm ∴OX=8cm Although, you are getting the right ans, here but the method is wrong OE - OX = XE 10 - 8 cm = 2 cm ∴ XE = 2 cm
2) In circle 'b' OP is radius PQ is chord O′X⊥PQ⇒PX=12PO→proved⇒PX=6cm→provedPO′−PX=12O′X 8 - 6 = 2 cm ∴O′X=4cmO′D−O′X=XD8−4=4cm∴XD=4cm
3) ∴OO′=(OX+O′X)−(DX+XE) =(8+4)-(2+4) =12-6 =6 cm
1) We cannot say that,OX = 8 cm Consider ΔOPX OP is radius
OX has to be calculated using pythagoras theorem. 102=62+OX2 ⇒OX2=100−36=64 ⇒OX=8cm
2) |||ly in ΔPO′X PO′2=PX2+XO′2⇒82=62+XO′2⇒XO′2=64−36=28⇒XO′=√28 ∴ Distance between centres