Question 1146382
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Clearly with a center (2,1) and radius 2, one of the two lines tangent to the circle passing through the origin is x=0.<br>
Some work is needed to find the other line....<br>
Let (a,b) be the other point of tangency to the given circle of a line that passes through the origin.  Then<br>
(1) The distance from (2,1) to (a,b) is 2:<br>
{{{(b-1)^2+(a-2)^2 = 2^2}}}
{{{b^2-2b+1+a^2-4a+4 = 4}}}
{{{a^2+b^2-4a-2b = -1}}}<br>
(2) The slope of the radius to the point of tangency is {{{(b-1)/(a-2)}}}
The slope of the tangent line is the negative reciprocal, {{{(2-a)/(b-1)}}}
The tangent passes through the points (0,0) and (a,b); so an equation of the tangent is
{{{b = ((2-a)/(b-1))a}}}
{{{b(b-1) = a(2-a)}}}
{{{b^2-b = 2a-a^2}}}
{{{a^2+b^2-2a-b = 0}}}<br>
Subtracting the equation in (1) from the equation in (2),<br>
{{{2a+b = 1}}}
{{{b = 1-2a}}}<br>
Substituting b=1-2a in (2)...<br>
{{{(1-2a)(-2a) = 2a-a^2}}}
{{{-2a+4a^2 = 2a-a^2}}}
{{{5a^2-4a = 0}}}
{{{a(5a-4) = 0}}}
{{{a = 4/5}}}<br>
And then, to find b<br>
{{{b = 1-2a = -3/5}}}<br>
The point (a,b) is (4/5,-3/5).<br>
Since the tangent line passes through the origin, the equation of the line is {{{y = (-3/4)x}}}<br>
A graph showing part of the lower half of the given circle and the second tangent line; the first tangent line is of course x=0, the y-axis:<br>
{{{graph(400,400,-1,3,-2,2,-sqrt(4-(x-2)^2)+1,(-3/4)x)}}}