Question

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

Answer 1

Radius of first circle $=\frac{50}{2}=25$ cm
Radius of second circle $=\frac{34}{2}=17$ cm

Segment $CD$ is the chord and $CD=30$ cm
$\Rightarrow CE=\frac{30}{2}=15$ cm

In $△ACE,$
By using Pythagoras theorem
$AE=\sqrt{A{C}^2-C{E}^2}$
$=\sqrt{{25}^2-{15}^2}$
$=20$ cm

Similarly, in $△BCE$
$BE=\sqrt{B{C}^2-C{E}^2}$
$=\sqrt{{17}^2-{15}^2}$
$=8$ cm

Therefore, the distance between their centres is $(20+8)=28$ cm.