document.write( "Question 203857: Please help me about this :
\n" ); document.write( "Find R so that the circle x^2+y^2=R^2 is tangent to the line x+2y=4.
\n" ); document.write( "Thanx:)
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Algebra.Com's Answer #153812 by Alan3354(69443)\"\" \"About 
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Find R so that the circle x^2+y^2=R^2 is tangent to the line x+2y=4.
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\n" ); document.write( "The center of the circle is the Origin. A line from the center to the tangent point will be perpendicular to the line x+2y=4. See that? Since it's perpendicular, its slope will be the negative inverse of the line, and it passes thru the origin.
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\n" ); document.write( "Find the slope of x+2y=4 by putting it into slope-intercept form (that means solve for y).
\n" ); document.write( "y = (-1/2)x + 2 so the slope, m, is -1/2.
\n" ); document.write( "The slope of the line from the center, the Origin, to the tangent point will have a slope of +2.
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\n" ); document.write( "Use y-y1 = m*(x-x1) to find the eqn of the radial line.
\n" ); document.write( "y = 2x since the point is (0,0)
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\n" ); document.write( "Now solve the pair of eqns to find the tangent point.
\n" ); document.write( "x + 2y = 4
\n" ); document.write( "y = 2x
\n" ); document.write( "x + 2*(2x) = 4
\n" ); document.write( "5x = 4
\n" ); document.write( "x = 0.8
\n" ); document.write( "y = 1.6
\n" ); document.write( "The tangent point is (0.8,1.6).
\n" ); document.write( "The radius, R, is the distance from the Origin to the point.
\n" ); document.write( "\"R%5E2+=+0.8%5E2+%2B+1.6%5E2\"
\n" ); document.write( "\"R%5E2+=+3.2\"
\n" ); document.write( "R = sqrt(3.2) = 4*sqrt(5)/5
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