Question

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

Answer 1

AC = Radius of first circle

Diameter of first circle = 50 cm

Radius of first circle AC = $\frac{50}{2}=25 \:cm$

AD is the radius of second circle

Radius of second circle AD = $\frac{34}{2}=17 \:cm$

AB = Chord = 30 cm

Theorem : If two circles intersect at two points their center lies on the perpendicular bisector of the common chord .

SO, AE = EB = $\frac{AB}{2} =\frac{30}{2}$ =15 cm

Now in ΔAEC

$Hypotenuse ^2= Perpendicular^2+Base^2\\\\AC^2= AE^2+CE^2\\\\25^2 = 15^2+CE^2\\\\25^2 -15^2=CE^2\\\\\sqrt{25^2 -15^2}=CE$

CE = 20

Now in ΔAED

$Hypotenuse ^2= Perpendicular^2+Base^2\\\\AD^2= AE^2+ED^2\\\\17^2 = 15^2+ED^2\\\\17^2 -15^2=ED^2\\\\\sqrt{17^2 -15^2}=ED$

ED = 8

Distance between their center= CE+ED = 20+8 = 28 cm

$\boxed{\bold{\therefore \:The \:distance \:between \:their \:center \:is \:28 \:cm.}}$

Answer 2

AC = Radius of first circle

Diameter of first circle = 50 cm

Radius of first circle AC =  

AD is the radius of second circle

Radius of second circle AD =  

AB = Chord = 30 cm

Theorem : If two circles intersect at two points their center lies on the perpendicular bisector of the common chord .

SO, AE = EB =  =15 cm

Now in ΔAEC

CE = 20

Now in ΔAED

ED = 8

Distance between their center= CE+ED = 20+8 = 28 cm