document.write( "Question 1087375: Show that the circles x^2 + y^2 - 16x - 20y + 115 = 0 and x^2 + y^2 + 8x - 10y + 5 = 0 are tangent and fine the point of tangency \n" ); document.write( "
Algebra.Com's Answer #701655 by Boreal(15235)\"\" \"About 
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x^2 + y^2 - 16x - 20y =-115
\n" ); document.write( "x^2-16x+64+y^2-20y+100=49, adding 164 to both sides
\n" ); document.write( "(x-8)^2+(y-10)^2=7^2
\n" ); document.write( "Circle one has a center of (8, 10) and radius 7.
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\n" ); document.write( "x^2 + y^2 + 8x - 10y + 5 = 0;
\n" ); document.write( "x^2+8x+16+y^2-10y+25=36
\n" ); document.write( "(x+4)^2+(y-5)^2=6^2
\n" ); document.write( "Circle two has a center of (-4, 5) and radius 6
\n" ); document.write( "The tangent line is perpendicular to the line connecting the two radii.
\n" ); document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "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.
\n" ); document.write( "y=5+(6/13)*5=(65/13)+(30/13)=95/13
\n" ); document.write( "the point is at (20/13, 95/13)
\n" ); document.write( "the line equation is y-(95/13)=(-12/5)(x-20/13)
\n" ); document.write( "This is y=(-12/5)x+(715/65), OR y=(-12/5)x+11\r
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\n" ); document.write( "\n" ); document.write( "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.\r
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