Question 969255
Write the equation of the circle satisfying the given condition.
Radius of  {{{2sqrt(5)}}}, tangent to y = 2x and passing through (3,-4)
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The center has to be on the line parallel to y = 2x and {{{2sqrt(5) = sqrt(20)}}} units from the line.
Find the line's equation:
y = 2x has a slope of 2 and passes thru (0,0)
A line perpendicular to it (RS) thru (0,0) is y = -x/2
A point on the line RS sqrt(20) from (0,0) is the intersection of a circle centered at (0,0) with r = sqrt(20) 
--> x^2 + y^2 = 20
y = -x/2
x^2 + x^2/4 = 20
x = 4  (Ignore the x = -4 in Q2, (3,-4) is in Q4)
--> (4,-2)
The center of the circle is on a line thru (4,-2) with m = 2 (parallel to y = 2x)
y+2 = 2(x-4) = 2x - 8
y = 2x - 10
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Find the point(s) sqrt(20) from (3,-4) on the line y = 2x - 10
It's the intersections of a circle centered at (3,-4) r = sqrt(20)
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{{{x-3)^2 + (y+4)^2 = 20}}}
Sub for y
{{{(x-3)^2 + (2x-6)^2 = 20}}}
{{{(x-3)^2 + 4(x-3)^2 = 20}}}
{{{(x-3)^2 = 4}}}
x = 5 --> center @ (5,0) --> {{{(x-5)^2 + y^2 = 20}}}  ** Circle #1
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x = 1 --> center @ (1,-8) --> {{{(x-1)^2 + (y+8)^2 = 20}}}  ** Circle #2
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There must be an easier way to do this, but I don't know of one.