How to calculate cross point $(x, y)$ of circle circumference and $90^\circ$ triangle

I got stuck on this problem, that is rather easy to present but I don't know how to solve it.

So I want to get $x$, $y$ coordinates of the point where circle and hypotenuse crosses. Circle radius is $2$ making shorter catet $2$ and longer $4$, hypotenuse $\sqrt{20}$.

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The coordinates of the circle center seem to be $(-1,0)$. Is it correct? –  Américo Tavares Mar 26 '12 at 21:05
yes, it was shifted on this example, could be 0, 0 too –  MarkokraM Mar 26 '12 at 21:07

I'm going to assume the circle is centered at the origin. You can correct the answer accordingly by shifting the coordinates in the solution. We can describe the circle with

$$x^{2} + y^{2} = 4$$

So that (for the upper-half semicircle)

$$y = \sqrt{4 - x^{2}}$$

Now, we find an equation for the line. We know that

$$m = \frac{\Delta y}{\Delta x} = \frac{4}{2} = 2$$

We also are given one point on the line: $(0,-2)$, so that the $y$-intercept is $-2$. Thus, the equation for the line is

$$y = 2x - 2$$

To find the intersection, we solve

$$\sqrt{4 - x^{2}} = 2x - 2$$

which gives $x = \frac{8}{5}$, and plugging this into $y = 2x - 2$, we obtain $y = \frac{6}{5}$

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All answers were good in my opinion. This one gave me more general and to my mind understood able solution. So I accept it as a correct answer. I also wanted to picture answer with extra application, pentagram based on r crossing original intersection point we found: desmond.imageshack.us/Himg29/… Thanks guys, good work, I learned/recalled a lot from college math :) –  MarkokraM Mar 28 '12 at 11:19

It looks like your circle is centered at $(-1,0)$. Because it is of radius $2$, the equation for the circle is $$(x+1)^2+y^2=4.$$ The bottom left corner of the triangle is at $(-1,-2)$. The slope of the hypotenuse is $2$, because the "rise / run" is equal to $4/2=2$. Therefore, the equation of the line that coincides with the hypotenuse is $$(y+2)=2(x+1)$$ which can be simplified to $$y=2x.$$ Thus, the hypotenuse intersects the circle when both of the equations $$(x+1)^2+y^2=4\qquad y=2x$$ are satisfied. We can find when that happens by substituting $2x$ for each occurrence of $y$ in the first equation: $$(x+1)^2+(2x)^2=x^2+2x+1+4x^2=5x^2+2x+1=4$$ Solving the quadratic equation $$5x^2+2x-3=0$$ we get that $$x=\frac{-2\pm\sqrt{4+4\cdot5\cdot3}}{10}=\frac{-2\pm 8}{10}=-1\text{ or }\frac{3}{5}.$$ Therefore, the hypotenuse intersects the circle at the point $(-1,-2)$ (this is the bottom left corner of the triangle) and at the point $(\frac{3}{5},\frac{6}{5})$ (this is the point you're interested in).

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I like result of this formula, but is it possible to set up starting point somewhere there (-1, -2)? I'm not sure how to use this with variables w (x1), h (y1), r (radius) –  MarkokraM Mar 26 '12 at 22:53
I can get right answer by this definition with values presented on the ordinal question, but when I change variables, results is not what I expect. Can you tell what is wrong with this: a = ((x-k)^2+(y-h)^2)-(r^2); b = (2*x)-h-y; Where k = x and h = y on plot and r = radius. –  MarkokraM Mar 27 '12 at 1:16

You have the circle $(x+1)^2+y^2=4$ and the line $y=2x$. Substitution to get rid of $y$ gives the equation \begin{align} 4&=(x+1)^2+4x^2\\ 0&=5x^2+2x-3 \end{align} Solving with the quadratic formula yields $x=-1$ and $x=\frac35$. Therefore, the point in question is $\left(\frac35,\frac65\right)$.

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