i have two circles with intersecting each other. two circles have same radius.they intersects in two points. i like find formula for finding that two points

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- Jan 20th 2010, 08:53 PM #1

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- Jan 21st 2010, 02:02 AM #2
If you have the circles in a Cartesian form, eg. $\displaystyle ax^2+by^2 = r^2$ and $\displaystyle cx^2+dy^2 = r^2$ where $\displaystyle r$ is the radius, you know that at the two intersecting points the coordinates of the two circles are the same.

Also you know that the radius are the same, therefore you can write it as:

$\displaystyle ax^2+by^2 = cx^2+dy^2$ and solve for $\displaystyle x,y$.

- Jan 21st 2010, 05:55 AM #3

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This is an incorrect formula. "$\displaystyle ax^2+ by^2= r^2$" is the equation of an

**ellipse**with center at the origin. Since you have two circles, you cannot assume they both have center at the origin so you are better off writing them as $\displaystyle (x-a)^2+ (y- b)^2= r^2$ and $\displaystyle (x- c)^2+ (y- d)^2= r^2$ (with same radius). You can solve those two equations for x and y, in terms of a, b, c, and d, to find the points at which they intersect.

But a more geometric method and, I think, simpler is this: Because the two circles have the same radius, the line between the two points of intersection is the perpendicular bisector of the line between the two centers. Knowing the coordinates of the centers, you can find the coordinates of the midpoint and the slope, m, of the line between centers. The perpendicular bisector is the line through that midpoint with slope -1. Find the points at which that line intersects either circle.

In fact, here is what I woud do: Given that one circle has center (x0,y0) and radius r and the other has center (x1, y1) and radius r, I would first**translate**so that the first circle has center at (0,0), by subtracting x0 from every x coordinate and y0 from every y coordinate. That means that the second circle has center at (a, b) where a= x1- x0 and b= y1- y0. The line between centers is now just y= (b/a)x, which has slope b/a, and the midpoint is (a/2, b/2). The perpendicular bisector is given by y= (-a/b)x. Also the first circle now has equation $\displaystyle x^2+ y^2= r^2$ so the line y= (-b/a)x will intersect the first circle where $\displaystyle x^2+ (b^2/a^2)x^2= x^2(1+ b^2/a^2)$$\displaystyle = \frac{a^2+ b^2}{a^2} x^2= r^2$. Now $\displaystyle x^2= \frac{a^2r^2}{a^2+ b^2}$ and $\displaystyle x= \pm\frac{ar}{\sqrt{a^2+ b^2}}$. Of course, y, then, is $\displaystyle y= (b/a)x= \pm\frac{br}{\sqrt{a^2+ b^2}}$, using the same sign as the corresponding y: $\displaystyle \left(\frac{ar}{\sqrt{a^2+ b^2}}, \frac{br}{\sqrt{b^2+ a^2}}\right)$ and $\displaystyle \left(\frac{-ar}{\sqrt{a^2+b^2}}, \frac{-br}{\sqrt{b^2+ a^2}}\right)$.

Finally, translate back by adding x0 to the x coordinate and y0 to the y coordinate.