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find the equation of the circle touching the line x+2y=8 at (0,4)and passing through (3,7).
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\n" ); document.write( "\"touching\" must mean tangent at that point, otherwise there's no single solution.
\n" ); document.write( "Since it's tangent, a diameter will be perpendicular to the line and pass thru (0,4). The slope of that line is +2, the negative inverse of the line x+2y=8.
\n" ); document.write( "The eqn of the line thru the center of the circle and thru (0,4) is:
\n" ); document.write( "y-4 = 2x
\n" ); document.write( "Now, the center of the circle is on that line and equidistant from (0,4) and (3,7).
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\n" ); document.write( "Any point (x,y) on line y = 2x+4 is, from (0,4) of sqrt[x^2 + (y-4)^2], and
\n" ); document.write( "from (3,7) sqrt[(y-7)^2 + (x-3)^2]. Note: (y-7) or (7-y) is the same, since it's squared.
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\n" ); document.write( "Set those equal:
\n" ); document.write( "sqrt[x^2 + (y-4)^2] = sqrt[(y-7)^2 + (x-3)^2]
\n" ); document.write( "x^2 + y^2-8y+16 = y^2-14y+49 + x^2-6x+9
\n" ); document.write( "-8y+16 = -14y+49 - 6x+9
\n" ); document.write( "6y = -6x + 42
\n" ); document.write( "y = -x + 7
\n" ); document.write( "This is the center, so it's on the line y = 2x+4
\n" ); document.write( "So -x + 7 = 2x+4
\n" ); document.write( "3x = 3
\n" ); document.write( "x = 1
\n" ); document.write( "y = 6
\n" ); document.write( "The center is (1,6)
\n" ); document.write( "To check:
\n" ); document.write( "(1 + 4) = 4 + 1 (squares of distances)
\n" ); document.write( "It is equidistant, and it's on the line, so it's correct.
\n" ); document.write( "Jeez\r
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