Question 1087375
{{{x^2-16x+y^2 - 20y + 115 = 0}}}
{{{(x-8)^2+(y-10)^2= -115+64+100 }}}
{{{(x-8)^2+(y-10)^2=49}}}
{{{(x-8)^2+(y-10)^2=7^2}}}
and
{{{x^2+8x + y^2 - 10y + 5 = 0}}} 
{{{(x+4)^2+(y-5)^2=-5+16+25}}}
{{{(x+4)^2+(y-5)^2=36}}}
{{{(x+4)^2+(y-5)^2=6^2}}}
So the point of tangency lies on the line that connects the centers of the circles (8,10) and (-4,5). 
If the distance from the centers is x then the point lies {{{(6/(6+7))}}} of the distance from (-4,5) to (8,10).
So the x distance from the centers is,
{{{dx=8-(-4)=12}}}
So then starting from -4,
{{{x[t]=-4+12(6/13)=-4+72/13=-52/13+72/13=20/13}}}
And the y distance is,
{{{dy=10-5=5}}}
And starting from 5,
{{{y[t]=5+5(6/13)=65/13+30/13=95/13}}}
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({{{20/13}}},{{{95/13}}})
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*[illustration fr8.JPG].