Question

The length of common chord of two intersecting circles is 30 cm. If the diameters of these two circles be 50 cm & 34 cm, the distance between their centres 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 in $\Delta$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.