prove that radius is a root of $(3-2\sqrt{2})t^2-2\sqrt{2}t+2=0$

Question

a circle is described to pass through the origin and to touch the lines $x=1,x+y=2$ then prove that the radius of the circle is a root of $(3-2\sqrt{2})t^2-2\sqrt{2}t+2=0$

solution i try

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length of perpendicular from $(1-r,2r-1)$ is $\displaystyle \frac{1-r+2r-1-2}{\sqrt{2}}=r^2$

$r=2(\sqrt{2}-1)$ which not satisfy $(3-2\sqrt{2})t^2-2\sqrt{2}t+2=0$

How to find correct radious in that question

It seems that you have a mistake: the coordinates of the center are $\alpha=1-r$ and $\beta=\pm\sqrt{2r-1}$. — Emilio Novati, Feb 18 '18 at 21:18

score 1 · Answer 1 · answered Feb 18 '18 at 08:55

1

let $$x_M,y_M$$ the coordinates of the centre of the circle, and $r$ the searched radius, then we get $$\left|\frac{x_M+y_M-2}{\sqrt{2}}\right|=r$$ $$|x_M-1|=r$$ $$x_M^2+y_M^2=r^2$$ can you finish?

answered Feb 18 '18 at 08:55

Dr. Sonnhard Graubner

95,283

prove that radius is a root of $(3-2\sqrt{2})t^2-2\sqrt{2}t+2=0$

1 Answers1