Geo4.77

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Show that the circle x^{2}+y^{2}-4x-6y-12=0,x^{2}+y^{2}+6x+18y+26=0\, touch each other and find the point of contact

Centres of the two circles is

A(2,3),B(-3,-9)\,

Radius is

r_{1}={\sqrt  {4+9+12}}=5,r_{2}={\sqrt  {9+81-26}}=8\,

Distance between the two centres

d={\sqrt  {(2+3)^{2}+(3+9)^{2}}}={\sqrt  {25+144}}={\sqrt  {169}}=13\,

Now

d=r_{1}+r_{2}\,. Hence the two circles touch eachother.

The equation of the line joing the two centres is

y-3={\frac  {-9-3}{-3-2}}(x-2)\,

5y-15=12x-24\,

12x-5y-9=0\,

C=point of contact of two circles.C divides Bin the ratio 5:8

Hence

C=\left({\frac  {5(-3)+8(2)}{5+8}},{\frac  {5(-9)+8(3)}{5+8}}\right)\,

C=\left({\frac  {1}{13}},{\frac  {-21}{13}}\right)\,


Main Page:Geometry:Circles