I'm looking for a bit of help with my project. At this point the context is a bit irrelevant, but I'm looking to find the intersection points of multiple circles.

There are $3$ circles, with the mid-point co-ordinates of $(0,0) , (2,4)$ , and $(4,0)$. Their radii are given in terms of an unknown variable '$a$'.

I have defined my 3 circles using the standard circle formula of $(x-h)^2 + (y-k)^2 = r^2$

Circle $A: x^2 + y^2 = (0.97 \times a)^2$

Circle $B: (x-2)^2 + (y-4)^2 = a^2$

Circle $C: (x-4)^2 + y^2 = (0.43 \times a)^2$

As an end result I would like the $6$ intersection points in Cartesian form, assuming that each circle intersects with each other circle $2$ times.

I have tried solving for 'a' but always end up with my answer in terms of both x and y and then get a bit lost as to where to go from there.

This is my first time seeking online help for project help before, so please be kind. Any advice or suggestions would be greatly appreciated.

Thankyou :)

Edit: Just to add some clarification, the fact that I'm working with a 3rd variable is what is throwing me off I think. Should the radii of these circles be known values, I would not have any issues with this solution.

  • $\begingroup$ look at this $\endgroup$
    – Nick
    Aug 13, 2016 at 5:52
  • $\begingroup$ $a$ is not a variable. It is a constant. Your answer will be in terms of $a$. Solve using that $a$ is just like any other numerical value. $\endgroup$
    – GoodDeeds
    Aug 13, 2016 at 6:04

2 Answers 2


let $$x^2+y^2-(\frac{97}{100}a)^2=0$$ the first and $$(x-2)^2+(y-4)^2-a^2=0$$ the second equation, then the difference is given by $$-20+\frac{591a^2}{100000}+4x+8y=0$$ solve this equation for $y$ plug this in one of the given equations above and you can calculate the $x$ coordinate


In this case, $a$ is not the same type of unknown quantity as $x$ and $y$ are. Rather, look at $a$ as an input parameter. What you are looking for are the $x$ and $y$ coordinates of the intersections in terms of $a$. Pretend that $a$ is just like any other number, and solve for $x$ and $y$. Said a different way, treat $a$ as a known quantity.

Let's just look at the intersections of circles A and B first. An intersection of two curves occurs when a point satisfies both of the equations. You have two equations and two unknowns, so you can use the elimination and substitution method to solve this system. It sounds like this is the part that you already understand, but I'll show my steps so that you can see how $a$ filters through.

1) Expand the equations $$\begin{alignat}{4} &x^2 +{} &&y^2 &&= 0.941&&a^2 \\ &x^2 -4x + 4 +{} &&y^2 - 8y + 16 &&= &&a^2 \\ \end{alignat} $$ 2) Subtract the second equation from the first to eliminate the $x^2$ and $y^2$ terms. $$ 4x - 4 + 8y -16 = -0.0591a^2 \\ $$ 3) Simplify and solve for $x$. Solving for $y$ would work too. $$ x = -2y - 0.0148a^2 + 5 $$ 4) Substitute this expression for $x$ back into one of the original equations for circles A or B. I'm using circle A and putting it into the standard quadratic form. $$ y^2 + y(0.0118a^2-4) + (0.0000437a^4 - 0.218a^2 + 5) = 0 $$ 5) As messy as this looks, it is just like any other quadratic equation. I will use the quadratic formula here rather than attempt factoring. The term $y_{AB}$ is the y coordinate of the intersections of circles A and B. $$ y_{AB} = \left\{\begin{array}{l} -0.00591a^2+2 + \sqrt{-0.00000873a^4 + 0.194a^2 - 1} \\ -0.00591a^2+2 - \sqrt{-0.00000873a^4 + 0.194a^2 - 1} \end{array}\right\} $$ 6) Substituting this back into the equation from step 3 gives the solutions for $x_{AB}$. $$ x_{AB} = \left\{\begin{array}{l} -0.00296a^2 + 1 - 2\sqrt{-0.00000873a^4 + 0.194a^2 - 1} \\ -0.00296a^2 + 1 + 2\sqrt{-0.00000873a^4 + 0.194a^2 - 1} \end{array}\right\} $$ It's worth mentioning significant digits here. I only bothered to type 3, but I carried full double precision values through the calculations and only rounded while typing. If you round at intermediate steps, your solution can sometimes lose more precision than you think.

Finding the intersections A, C and B, C work exactly the same way.

The minimum value of $a$ that produces all six intersections is 3.13, and the maximum value is 7.41.

  • $\begingroup$ Hi James, Thankyou so much for your assistance thus far. Can I just ask how you went from having the two x,y points defined above to producing possible values for 'a' like 3.13? $\endgroup$
    – Matt
    Aug 14, 2016 at 21:36
  • $\begingroup$ @Matt, Sure. If you look at the expressions under the square root signs, they determine how many times two circles intersect. (This is the "determinant" of the quadratic equation). Positive = two intersections, zero = one intersection, negative = no (real) intersections. Just set each of the three determinants to zero and solve for $a$.Since circle C is the smallest and it is a little bit farther away, it is the limiting factor. Plotting definitely helps to visualize what is going on. $\endgroup$ Aug 15, 2016 at 14:50

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