Question

The length of the common chord of two intersecting circles is 30 cm. If the diameters of these two circles are 50 cm & 34 cm, the distance between their centres is ____ cm.

• A. 25 cm
• B. 28 cm
• C. 30 cm
• D. 52 cm

Answer 1

The correct option is B 28 cm

The perpendicular drawn from centre of the circle bisects the chord.
OO' bisects the chord AB into two equal parts of 15 cm each.
$AC=CB=15cm$
Now applying Pythagoras theorem to $\Delta$ OAC, we get
$(25{)}^2=(15{)}^2+(x{)}^2$
$625=225+(x{)}^2$
$\Rightarrow x=20$ cm
Now, applying Pythagoras theorem in $\Delta$O'AC we get,
$(17{)}^2=(15{)}^2+(y{)}^2$
$289=225+(y{)}^2$
$y=8$ cm
Therefore distance between the two centres is $(x+y)=(20+8)=28$ cm.