document.write( "Question 1209461: The line y= mx + c is tangent to the curve x^2 + y^2 = 4. Prove that 4m^2 = c^2 - 4 \n" ); document.write( "
Algebra.Com's Answer #848883 by math_tutor2020(3817)\"\" \"About 
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\n" ); document.write( "x^2 + y^2 = 4
\n" ); document.write( "x^2 + (mx+c)^2 = 4 ................. plug in y = mx+c
\n" ); document.write( "x^2 + m^2x^2+2mcx+c^2 = 4
\n" ); document.write( "(1+m^2)x^2 + 2mcx + c^2-4 = 0\r
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\n" ); document.write( "\n" ); document.write( "We have the quadratic function
\n" ); document.write( "f(x) = (1+m^2)x^2 + 2mcx + (c^2-4)
\n" ); document.write( "where,
  • x^2 coefficient = (1+m^2)
  • x coefficient = 2mc
  • constant = (c^2-4)
If the discriminant is 0, then the quadratic has exactly one root.
\n" ); document.write( "This generates exactly one point of intersection between the circle and the tangent line.\r
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\n" ); document.write( "\n" ); document.write( "d = discriminant
\n" ); document.write( "d = (x coefficient)^2 - 4*(x^2 coefficient)*(constant)
\n" ); document.write( "d = (2mc)^2 - 4*(1+m^2)(c^2-4)
\n" ); document.write( "(2mc)^2 - 4*(1+m^2)(c^2-4) = 0
\n" ); document.write( "4m^2c^2 - 4*(1+m^2)(c^2-4) = 0
\n" ); document.write( "4( m^2c^2 - (1+m^2)(c^2-4) ) = 0
\n" ); document.write( "m^2c^2 - (1+m^2)(c^2-4) = 0/4
\n" ); document.write( "m^2c^2 - (c^2-4) - m^2(c^2-4) = 0
\n" ); document.write( "m^2c^2 - (c^2-4) - m^2c^2+4m^2 = 0
\n" ); document.write( "-(c^2-4)+4m^2 = 0
\n" ); document.write( "4m^2 = c^2-4
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