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 centers is ____ cm.

• A. 30
• B. 28
• C. 24
• D. 32

Answer 1

The correct option is B

28

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

${\left(25\right)}^2$ = ${\left(15\right)}^2$ + ${\left(x\right)}^2$

625 = 225 + ${\left(x\right)}^2$

$⟹$ $x$ = 20 cm

Now, applying Pythagoras theorem in $\Delta$O'AC we get,

${\left(17\right)}^2$ = ${\left(15\right)}^2$ + ${\left(y\right)}^2$

289 = 225 + ${\left(y\right)}^2$

y = 8 cm

Therefore distance between the two centres is $(x+y)=(20+8)=28$ cm.