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

enter image description here

So I want to get x,y coordinates of the point where circle and hypothenuse crosses. Circle radius is 2 making shorter catet 2 and longer 4, hypothenuse square root of 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 –  PHPGAE Mar 26 '12 at 21:07
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3 Answers 3

up vote 1 down vote accepted

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 :) –  PHPGAE Mar 28 '12 at 11:19
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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) –  PHPGAE 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. –  PHPGAE Mar 27 '12 at 1:16
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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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