Question

Two circles of radii 8 cm and 3 cm have a direct common tangent of length 10 cm. Find the
distance between their centres, upto two places of decimal.
part.​

Answer 1

• The distance between those centres = 11.18 cm.

Diagram :

$\setlength{\unitlength}{2mm}\begin{picture}(0,0)\put(0,0){\thicklines\circle{10}}\put(20,0){\thicklines\circle{6.9}}\put(0,0){\thicklines\line(3,0){4cm}}\put(0,0){\thicklines\line(0,3){7mm}}\put(0,3.5){\thicklines\line(5,0){4cm}}\put(20,0){\thicklines\line(0,3){7mm}}\put(-1,-1){\sf\footnotesize A}\put(-1,4){\sf\footnotesize B}\put(20,4){\sf\footnotesize C}\put(20,-1){\sf\footnotesize D}\put(0.7,3.4){\line(0,-3){1.3mm}}\put(0,2.7){\line(3,0){1.5mm}}\put(20,2.7){\line(-3,0){1.5mm}}\put(19.3,3.5){\line(0,-3){1.5mm}}\put(-2.8,1.5){\sf\footnotesize 8cm}\put(8,5){\sf\footnotesize 10cm}\put(20.3,1.5){\sf\footnotesize 3cm}\end{picture}$

Given :

Radius of two circles :

1. The radius of the first circle (AB) = 8 cm.
2. The radius of the second circle (CD) = 3 cm.

• The length of a direct common tangent (BC) = 10 cm.

To Find :

• The distance between their centres.

Solution :

We have to find the distance between their centres (AD).

We know that,

$\boxed{ \rm{ \green{length \: of \: direct \: tangent = \sqrt{ {(d)}^{2} - {( r_{1} - r_{2})}^{2} } }}}$

We have,

$\bullet \: \: \rm length \: of \: direct \: tangent = BC = 10 \: cm \\ \\ \bullet \: \: \rm r_{1} = AB = 8 \: cm \\ \\ \bullet \: \: \rm r_{2} = CD = 3 \: cm \\ \\ \bullet \: \: \rm d = AD = \: ?$

Now, substitute all the given values in the formula of "Length of direct tangent".

$:\implies \rm BC = \sqrt{ {AD}^{2} - {(AB - CD)}^{2} } \\ \\ : \implies \rm 10 \: cm = \sqrt{ {AD}^{2} - {(8 \: cm - 3 \: cm)}^{2}} \\ \\ : \implies \rm 10 \: cm = \sqrt{ {AD}^{2} - {(5 \: cm)}^{2} } \\ \\ : \implies \rm {(10 \: cm)}^{2} = {AD}^{2} - 25 \: {cm}^{2} \\ \\ : \implies \rm100 \: {cm}^{2} = {AD}^{2} - {25 \: cm}^{2} \\ \\ : \implies \rm{ 100 \: cm}^{2} + {25 \: cm}^{2} = {AD}^{2} \\ \\ : \implies \rm125 \: {cm}^{2} = {AD}^{2} \\ \\ : \implies \rm \sqrt{125 \: {cm}^{2} } = AD \\ \\ : \implies \rm\sqrt{5 \times 5 \times 5 \: {cm}^{2} } = AD \\ \\ : \implies \rm5 \sqrt{5} \: cm = AD \\ \\ : \implies \rm11.18 \: cm = AD \\ \\ \: \therefore \: \: \rm AD = 11.18 \: cm$

Hence,

The distance between the centres of two circles is 11.18 cm.