Question 1087375
 x^2 + y^2 - 16x - 20y =-115
x^2-16x+64+y^2-20y+100=49, adding 164 to both sides
(x-8)^2+(y-10)^2=7^2
Circle one has a center of (8, 10) and radius 7.
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x^2 + y^2 + 8x - 10y + 5 = 0;
x^2+8x+16+y^2-10y+25=36
(x+4)^2+(y-5)^2=6^2
Circle two has a center of (-4, 5) and radius 6
The tangent line is perpendicular to the line connecting the two radii.
That line has slope 5/12 and its formula is y-y1=m(x-x1); m slope (x1,y1) point. y-10=(5/12)(x-8), or y=(5/12)x+80/12. The slope of the tangent line is -12/5, the negative reciprocal.

The distance between the two centers is 13, so the x component of the tangent point is 6/13 the way from -4 to 8, which has distance 12.  That is x=-4+(6/13)*12=-4+72/13, or 20/13.
y=5+(6/13)*5=(65/13)+(30/13)=95/13
the point is at (20/13, 95/13)
the line equation is y-(95/13)=(-12/5)(x-20/13)
This is y=(-12/5)x+(715/65), OR y=(-12/5)x+11

{{{graph(300,300,-15,15,-15,15,10+sqrt(49-(x-8)^2),10-sqrt(49-(x-8)^2),5+sqrt(36-(x+4)^2),5-sqrt(36-(x+4)^2),(-12/5)x+11,(5/12)x+80/12)}}}

If you set the two circle equations equal to each other, the square terms cancel each other and the result is 24x+10y=110 or 12x+5y=55.  The point (20/13, 95/13) is a solution to that equation.