Geo4.95

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Find the equation of the circle whose radius is 5units and which touches the circle x^{2}+y^{2}-2x-4y-20=0\, at the point (5,5).

equation of the tangent at(5,5) to the circle is

x(5)+y(5)-(x+5)-2(y+5)-20=0\,

4x+3y-35=0\,

Equation of the circle touching the the circle at (5,5) is

x^{2}+y^{2}-2x-4y-20+k(4x+3y-35)=0\,

x^{2}+y^{2}-2x(1-2k)-y(4-3k)-20-5k=0\,

Given radius is 5.

Hence

4(1-2k)^{2}+(4-3k)^{2}+4(20+5k)=4(25)\,

4(1+4k^{2}-4k)+16+9k^{2}-24k+80+20k=100\,

k^{2}+4k=0,k=0,-4\,

Therefore,the required circle is

x^{2}+y^{2}-2x-4y-20-4(4x+3y-35)=0\,

x^{2}+y^{2}-18x-16y+120=0\,

Main Page:Geometry:Circles