document.write( "Question 194599: give the equation of a circle tangent to the line 3x+y-2=0 at (-1,5) and with radius of 3.16227766 \n" ); document.write( "
Algebra.Com's Answer #148414 by Alan3354(69443)\"\" \"About 
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give the equation of a circle tangent to the line 3x+y-2=0 at (-1,5) and with radius of 3.16227766
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\n" ); document.write( "There are 2 circles, one on either side of the line.
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\n" ); document.write( "3x+y-2=0 has a slope, m, of -3, so the line thru (-1,5) and the center has a slope of +1/3.
\n" ); document.write( "The center is a distance of sqrt(10) from (-1,5)
\n" ); document.write( "The eqn of the line thru the center and the point (-1,5) is found by:
\n" ); document.write( "y-y1 = m*(x-x1) where (x1,y1) is the point (-1,5)
\n" ); document.write( "y-5 = (1/3)*(x+1)
\n" ); document.write( "y = x/3 + 16/3
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\n" ); document.write( "The center is sqrt(10) units from (-1,5) along this line.
\n" ); document.write( "Using Pythagoras:
\n" ); document.write( "10 = (x+1)^2 + (y-5)^2
\n" ); document.write( "Sub for y
\n" ); document.write( "10 = (x+1)^2 + (((x+16)/3) - 5)^2)
\n" ); document.write( "10 = x^2 + 2x + 1 + (1/9)*(x^2 + 2x + 1)
\n" ); document.write( "90 = 9x^2 + 18x + 9 + x^2 + 2x + 1
\n" ); document.write( "90 = 10x^2 + 20x + 10
\n" ); document.write( "x^2 + 2x - 8 = 0
\n" ); document.write( "x = -4, x = 2
\n" ); document.write( "Centers at (-4,4) and at (2,6)
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\n" ); document.write( "Center at (-4,4): (x+4)^2 + (y-4)^2 = 10
\n" ); document.write( "Center at (2,6): (x-2)^2 + (y-6)^2 = 10\r
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