The circle $$x^2 +y^2 -4x-4y+4=0$$ is inscribed in a triangle which has two of its sides along the coordinate axes. How can we find the equation of the third line.

Now I have assumed that the line's equation would be

$${x \over a} + { y \over b} =1$$ where $a$ serves as the x intercept and $b$ as the y intercept.

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Now since this line would be a tangent the perpendicular distance of the line from the centre of the circle would be equal to the radius of the circle.

Perpendicular distance from $(2,2)$ to $bx+ay=ab$:

$${|2b+2a-ab| \over \sqrt{b^2 +a^2}} = 2$$

This gives me one equation with two variables. How can I find other equations in order to solve this system? ANY HINT WOULD WORK.

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    $\begingroup$ There are infinitely such lines. This is the only equation possible. You can see this geometrically by creating some tangents of this circle. The tangents will meet both axes sometime $\endgroup$ – N.S.JOHN Oct 11 '16 at 8:03
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    $\begingroup$ @N.S. John. Not all tangents will meet with both axes. e.g. x= 4 or y=4. Also for certain points the tangents will not cross the positive axes. $\endgroup$ – MrYouMath Oct 11 '16 at 8:49
  • $\begingroup$ @MrYouMath sure thanks for adding that $\endgroup$ – N.S.JOHN Oct 11 '16 at 12:29

As N.S.JOHN points out in his comment, there are an infinite number of such lines. If you parameterize the circle as $x=2+2\cos t$, $y=2+2\sin t$, then $\mathbf n=(\cos t,\sin t)$ is a normal to the circle at $(x(t),y(t))$, and an equation of the tangent at this point is $$\mathbf n\cdot(\mathbf x-(2,2))=(x-2)\cos t+(y-2)\sin t=2.$$ This line, together with the coordinate axes, will form a circumscribed triangle for $0\lt t\lt\pi/2$.


That's the only equation you can get, because there are infinitely many lines that are possible. The only other condition is that the line must be a descending line in the first quadrant. In other words, $a,b>0$.

You can see that for any $a>4$, you can find a $b>0$ such that the condition will be satisfied.


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