Question 203857
Find R so that the circle x^2+y^2=R^2 is tangent to the line x+2y=4.
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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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Find the slope of x+2y=4 by putting it into slope-intercept form (that means solve for y).
y = (-1/2)x + 2 so the slope, m, is -1/2.
The slope of the line from the center, the Origin, to the tangent point will have a slope of +2.
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Use y-y1 = m*(x-x1) to find the eqn of the radial line.
y = 2x since the point is (0,0)
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Now solve the pair of eqns to find the tangent point.
x + 2y = 4
y = 2x
x + 2*(2x) = 4
5x = 4
x = 0.8
y = 1.6
The tangent point is (0.8,1.6).
The radius, R, is the distance from the Origin to the point.
{{{R^2 = 0.8^2 + 1.6^2}}}
{{{R^2 = 3.2}}}
R = sqrt(3.2) = 4*sqrt(5)/5