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Find the equation of the tangent at the point (0,3) on a circle of x sqr + y sqr = 1.
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\n" ); document.write( "Use ^ (Shift 6) for exponents.
\n" ); document.write( "x^2 + y^2 = 1
\n" ); document.write( "(0,3) is not on the circle.
\n" ); document.write( "If you mean thru (0,3) and tangent to the circle:
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\n" ); document.write( "(0,3) is P and the tangent point is T
\n" ); document.write( "O is the origin
\n" ); document.write( "OTP is a right triangle with T = 90 degs
\n" ); document.write( "side TP = sqrt(3^2 - 1^2) = sqrt(8)
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\n" ); document.write( "The tangent point is on the given circle and on the circle centered at (0,3) with radius sqrt(8) --> x^2 + (y-3)^2 = 8
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\n" ); document.write( "Find the intersections of the 2 circles (2 of them)
\n" ); document.write( "x^2 + (y-3)^2 = 8
\n" ); document.write( "x^2 + y^2 = 1
\n" ); document.write( "----------------------- Subtract
\n" ); document.write( "(y-3)^2 - y^2 = 7
\n" ); document.write( "y^2 - 6y + 9 - y^2 = 7
\n" ); document.write( "-6y = -2
\n" ); document.write( "y = 1/3
\n" ); document.write( "-------
\n" ); document.write( "x = -sqrt(8)/3 --> (-sqrt(8)/3,1/3)
\n" ); document.write( "x = +sqrt(8)/3 --> (+sqrt(8)/3,1/3)
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\n" ); document.write( "Find the eqn of the line thru the pairs of points
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\n" ); document.write( "|x y 1|
\n" ); document.write( "|0 3 1| = 0
\n" ); document.write( "|sq 1/3 1|
\n" ); document.write( "sq is the point (sqrt(8)/3,1/3)
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\n" ); document.write( "In Quadrant 1:
\n" ); document.write( "x*(3 - 1/3) - y*(0 - sqrt(8)) + (0 - sqrt(8)) = 0
\n" ); document.write( "8x/3 + sqrt(8)y = sqrt(8)
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\n" ); document.write( "You can find the other line. \n" ); document.write( "