If I draw two circles of radius 3 cm and 5cm with common centre and draw a line AB such that it is a chord to both the circles and length of CD is 2√5 cm, then find the distance of the chords from the centre and the length AC.
2 cm, 2.346 cm
Given, CD = 2√5 cm
Since OE⊥CD, CE=DE=√5 cm.
(Perpendicular from the centre to a chord bisects the chord.)
Apply Pythagoras' theorem in ΔOEC. We get
OE2+CE2=OC2
OE2+(√5)2=9
⟹OE2=9−5
∴OE2=4
⇒OE=2
In ΔOEA,
OE2+AE2=OA2
⟹22+AE2=52
⟹AE=√21 cm
Now,
AC=AE−CE =√21−√5
=4.582 cm−2.236 cm
=2.346 cm