How to find value of intersection point of two circles?

Question

Consider the following diagram

Here, we have a small circle of radius $r$ inside a large circle of radius $R$. The center of the small circle is offset to the right of the center of the large circle by some length $d$, such that both circles touch each other at one point.

Draw a line from this intersection point to a vertical from the center of the large circle, and call its length $x$.

How do we find $x$ when

1. $R=11,\, r=3, \, d=6$?
2. In general?

---

I started off with coordinate bashing, with $y=-\sqrt{R^{2}-x^{2}}$ as the equation for the large circle and $y=a-\sqrt{-x^{2}++2dx+r^{2}-d^{2}}$ as the equation for the smaller circle for some value $a$.

Then, I set them equal to one another, $$-\sqrt{R^{2}-x^{2}}=a-\sqrt{-x^{2}+2dx+r^{2}-d^{2}}$$ It quickly got very messy so I plugged in some numbers and got $$-\sqrt{121-x^2}=a-\sqrt{-x^2+12x-27}$$ Squaring twice with some manipulation gives $$-4 a^2 x^2+48 a^2 x-108 a^2=21904-296 a^2+a^4+x \left(24 a^2-3552\right)+144 x^2$$ which seems solvable but got wayyy too messy so I just wolfram alpha'd it to get

Setting both solutions equal to each other gives me $$-a^6+188 a^4-4480 a^2=0 \iff a^2 (a^2-160) (a^2-28)=0$$ giving me 5 solutions $a = 0, \pm 4\sqrt{10}, \pm 2\sqrt7$. Graphically, I see that $-2\sqrt{7}$ is the solution I want, and plugging it back into the monstrous expression of $x$ gives me $33\over4$ as the $x$ value.

---

Clearly, this method is extremely messy, and I imagine it would be even more messier for the general case. I suspect I am missing a simple geometric solution.

Any help?

Thanks! Max0815

Answer 1

From $OP = AT = x$ and $OT = R$ you get $PT=\sqrt{R^2-x^2}$.

From $OQ = d$ and $OM = R-r$ you get $QM = \sqrt{(R-r)^2-d^2}$.

Further consider the similarity of the triangles $OQM$ and $OPT$, leading to $$\frac{PT}{QM}=\frac{x}{d}\ .$$

Putting in the above (and squaring) then results in $$d^2 (R^2-x^2) = ((R-r)^2-d^2) x^2$$ or $$d^2 R^2 - d^2 x^2 = (R-r)^2 x^2 - d^2 x^2$$ which clearly results in $$x^2 = \frac{d^2 R^2}{(R-r)^2} = \left(\frac{d R}{R-r}\right)^2$$

--- rk

Answer 2

Since the little circle just touches the big circle at $D$ it is perpendicular to the tangent at that point. So if you extend the line to the back then you will meet the center $O$(because the line perpendicular to the tangent is the radius).

Let $\angle ODP = \theta$. Then $\angle ODP = \angle COQ = \theta$.

If we consider the triangle $COQ$ we can write $\displaystyle \cos\theta = \frac{OQ}{OC} = \frac{d}{R-r}$. Then we consider the triangle $OPD$ to write $\displaystyle x = OD\cos\theta = \frac{Rd}{R-r}$.

Substituting $R = 11, r = 3, d = 6$ to the above equation gives $x = \frac{33}{4}$.

Answer 3

$d+r\le R$, also $r\lt R$

$\displaystyle x=d+r\frac{d}{R-r}$

My reasoning? $x$ is the sum of $d$ and $r$ projected to the x-axis. The ratio of $d$ and $(R-r)$ defines the direction center of the big circle to tangent point of the two circles. The rest is Intercept theorem.
With the given values I get $\displaystyle x=\frac{33}{4}$

Answer 4

Let center of big circle be $O(0,0)$ and small circle be $o(d,-\sqrt{(R-r)^2-d^2})$. Then the equation of the line $\stackrel{\leftrightarrow}{Oo}$ is $$y=-\frac{\sqrt{(R-r)^2-d^2}}d x$$and it cuts the big circle $$x^2+y^2=R^2$$ at the tangency point $P$ with $x$-coordinate $$x=\frac{dR}{R-r}.$$