Two circles of radii 18 cm and 16 cm intersect each other and the length of their common chord is 20 cm. What is the distance (in cm) between their centres?

$4 \sqrt{14}+2 \sqrt{39}$

Here,

AB is common chord and C1C2 is the distance between the centres of the circle.

AB is being perpendicularly bisected at point D by C1C2.

\(\Delta \)AC1D & \(\Delta \)AC2D are right angled triangles.

AC1 = 18 cm

AC2 = 16 cm

AB = 20 cm

AD = \(\frac{20}{2}\) = 10 cm

From \(\Delta \)Ac1D,

\( { (AC2)}^{2 } \) = \( { (C2D)}^{2 } \) + \( { (AD)}^{2 } \)

= \( { 16}^{2 } \) = \( { (C2D)}^{2 } \) + \( { (10)}^{2 } \)

= \( { (C2D)}^{2 } \) = 256 - 100

= \( { (C2D)}^{2 } \) = 156

= \( { (C2D)}^{2 } \) = 4 x 39

= \( { (C2D)}^{2 } \) = 2\(\sqrt { 39}\)

Now, the distance two centres of the intersecting circles

= C1C2 = C1D + C2D = (4\(\sqrt { 14}\) + 2\(\sqrt { 39}\)) cm

Therefore, the distance between their centres is (4\(\sqrt { 14}\) + 2\(\sqrt { 39}\)) cm.