document.write( "Question 913984: Please help me solve this problem:\r
\n" ); document.write( "\n" ); document.write( "A circle of radius 1 is centered at the origin and passes through the point (1/2,3√/2).
\n" ); document.write( "(a) Find an equation for the line through the origin and the point (1/2,3√/2).
\n" ); document.write( "y= \r
\n" ); document.write( "\n" ); document.write( "(b) Find an equation for the tangent line to the circle at (1/2,3√/2).
\n" ); document.write( "y=
\n" ); document.write( "[Hint: A tangent line is perpendicular to the radius at the point of tangency.]
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Algebra.Com's Answer #554837 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
A circle of radius and centered at the origin and passes through the point \"P%281%2F2%2Csqrt%283%29%2F2%29\" .
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\n" ); document.write( "The equation for a line that starts with that starts with \"y =\" is the equation in slope-intercept form,
\n" ); document.write( "\"y=m%2Ax%2Bb\", where \"m\" is the slope and \"b\" is the y-intercept.
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\n" ); document.write( "(a) The line through the origin, \"O%280%2C0%29\" , and the point \"P%281%2F2%2Csqrt%283%29%2F2%29\"
\n" ); document.write( "has a y-intercept of \"b=0\" , and
\n" ); document.write( "a slope of \r
\n" ); document.write( "\n" ); document.write( "The equation for the line hrough the origin and the point \"P%281%2F2%2Csqrt%283%29%2F2%29\" that starts with \"y =\" is
\n" ); document.write( "\"highlight%28y=+sqrt%283%29x%29\" .
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\n" ); document.write( "(b) The tangent line to the circle at the point \"P%281%2F2%2Csqrt%283%29%2F2%29\" goes through the point of tangency, \"P%281%2F2%2Csqrt%283%29%2F2%29\" .
\n" ); document.write( "The radius of the circle at the point of tangency is the line segment that connects the points \"O%280%2C0%29\" and \"P%281%2F2%2Csqrt%283%29%2F2%29\" .
\n" ); document.write( "That line segment is part of the line with equation \"y=+sqrt%283%29x\" and slope \"m=sqrt%283%29\" found in part a).
\n" ); document.write( "According to the hint, the tangent you are looking for is perpendicular to that line and passes through the point of tangency, \"P%281%2F2%2Csqrt%283%29%2F2%29\" .
\n" ); document.write( "The line perpendicular to a line with slope \"m\", has a slope of \"-1%2Fm\" .
\n" ); document.write( "So, for the tangent you are looking for,
\n" ); document.write( "the slope is \"-1%2Fsqrt%283%29\" .
\n" ); document.write( "That is more elegantly expressed as
\n" ); document.write( "\"%28-1%2Fsqrt%283%29%29%2A%28sqrt%283%29%2Fsqrt%283%29%29=-sqrt%283%29%2F3\"
\n" ); document.write( "The equation of that tangent in point-slope form, based on point \"P%281%2F2%2Csqrt%283%29%2F2%29\" is
\n" ); document.write( "\"highlight%28y-sqrt%283%29%2F2=%28-sqrt%283%29%2F3%29%28x-1%2F2%29%29\" .
\n" ); document.write( "If you wanted to express that equation in slope intercept form, you would \"distribute\" and \"solve for y\", like this:
\n" ); document.write( "\"y-sqrt%283%29%2F2=%28-sqrt%283%29%2F3%29%28x-1%2F2%29\"--->\"y-sqrt%283%29%2F2=%28-sqrt%283%29%2F3%29x-%281%2F2%29%28-sqrt%283%29%2F3%29\"--->\"y-sqrt%283%29%2F2=%28-sqrt%283%29%2F3%29x%2Bsqrt%283%29%2F6\"--->\"y=%28-sqrt%283%29%2F3%29x%2Bsqrt%283%29%2F6%2Bsqrt%283%29%2F2\"--->\"y=%28-sqrt%283%29%2F3%29x%2Bsqrt%283%29%2F6%2B3sqrt%283%29%2F6\"--->\"y=%28-sqrt%283%29%2F3%29x%2B4sqrt%283%29%2F6\"--->\"highlight%28y=%28-sqrt%283%29%2F3%29x%2B2sqrt%283%29%2F3%29\"
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