Geo4.80

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Find the equation of the pair of direct common tangents to the following circles. x^{2}+y^{2}=16,x^{2}+y^{2}-4x-2y-4=0\,

Centre and radii to the given circles is

A=(0,0),r_{1}=4,B=(2,1),r_{2}={\sqrt  {4+1+4}}=3\,

AB={\sqrt  {(0-2)^{2}+(0-1)^{2}}}={\sqrt  {5}},r_{1}+r_{2}=7\,

AB<(r_{1}+r_{2})\,

These two circles intersect eachother.

Let C be the centre of similitude. C divides the circles in the ratio 4:3

Hence

C=\left({\frac  {4\cdot 2+3\cdot 0}{4+3}},{\frac  {4\cdot 1+3\cdot 0}{4+3}}\right)\,

C=\left({\frac  {8}{7}},{\frac  {4}{7}}\right)\,

Now the equation of the direct common tangents is

(x({\frac  {8}{7}})+y({\frac  {4}{7}})-16)^{2}=(x^{2}+y^{2}-16)({\frac  {64}{49}}+{\frac  {16}{49}}-16)\,

16(2x+y-28)^{2}=-704(x^{2}+y^{2}-16)\,

4x^{2}+y^{2}+784+4xy-56y-112x=-44x^{2}-44y^{2}+704\,

48x^{2}+45y^{2}+4xy-112x-56y+80=0\,

48x^{2}+4xy+45y^{2}-112x-56y+80=0\,


Main Page:Geometry:Circles