Question

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

Answer 1

Centre $A$ circle radius is $=\frac{50}{2}=25cm$

Centre $B$ radius is $=\frac{34}{2}=17cm$

Segment $DS$ is the chord

$DS=30cm$ (given)

As $DM=\frac{DS}{2}=15cm$

In $△AMD$, By using Pythagoras theorem

$A{D}^2=A{M}^2+M{D}^2$

$A{M}^2=A{D}^2-M{D}^2$

$A{M}^2=(25{)}^2-(15{)}^2$

$A{M}^2=400$

$AM=20cm$

Now in $△BMD$ using Pythagoras

$B{D}^2=D{M}^2+M{B}^2$

$M{B}^2=B{D}^2-D{M}^2\Rightarrow {17}^2-{15}^2$

$M{B}^2=64$

$MB=8cm$

In distance between the centres $AB$ which is $(AM+MB)=(20+8)=28cm$.