How to find the singular coordinate at which two circles touch

Question

I have two circles.

Circle 1 has the equation x^(2)+y^(2)=56900.

Circle 2 has the equation x^(2)-309.09090909...x+y^(2)-200.90909090...y=-31036.36363.

The ... is to represent that the decimal numbers are recurring.

I know that these two circles touch at a singular point, (200, 130) as shown in the below image.

Circles touching

But I can't figure out how to find that coordinate from the two circle equations. When I do attempt it I end up running into problems with complex numbers.

If someone could walk me through the steps to find the point where these circles touch (200,130) from the above equations it would be greatly appreciated.

Answer 1

The equations of the two circles are

$ x^2 + y^2 = 56900 \hspace{15pt} (1)$

and

$ x^2 - 309 \frac{1}{11} x + y^2 - 200 \frac{10}{11} y = - 31036 \frac{4}{11} \hspace{15pt} (2) $

Multiplying the second equation by $11$, gives

$ 11 x^2 - 3400 x + 11 y^2 - 2210 y = -341400 \hspace{15pt} (2') $

Multiplying the very first equation by $11$ gives

$ 11 x^2 + 11 y^2 = 625900 \hspace{15pt} (1') $

Now subtract equation $(2')$ from equation $(1')$,

$ 3400 x + 2210 y = 967300 $

Divide through by $170$

$ 20 x + 13 y = 5690 $

So that

$ y = \dfrac{1}{13} ( 5690 - 20 x) \hspace{15pt} (3) $

Substitute this into equation $(1)$

$ x^2 + \dfrac{1}{169} (5690 - 20 x)^2 = 56900 $

Expand and multiply through by $169$

$ 169 x^2 + 32376100 - 227600 x + 400 x^2 = 9616100 $

Simplify,

$ 569 x^2 - 227600 x + 22760000 = 0 $

Discriminant of this quadratic equation is

$ D = 227600^2 - 4 (569)(22760000) = 0 $

Since the discriminant is $0$, then there is only one solution given by

$ x = - \dfrac{B}{2 A} = - \dfrac{ - 227600}{2(569) } = 200 $

Substituting this into equation $(3)$, gives

$ y = \dfrac{1}{13} (5690 - 4000) = \dfrac{1690}{13} = 130 $

So that $(200, 130)$ is the only intersection point between the two circles.

Answer 2

If the radius of the big circle is $R$, and the radius of the small circle is $r$, and the center of the small circle is at $(a,b)$, then the location of the tangent point is

$$ \large \pmatrix{x\\y} = \pmatrix{ \frac{R\; a}{R-r} \\ \frac{R\; b}{R -r} } $$

Now this assumes you know the circles are tangent. So now the question becomes how to you extract the radius and center of a circle from its equation.

All you need to do and bring the equation into the following form

$$ (x-a)^2 + (y-b)^2 = r^2 $$ where $(a,b)$ is the center and $r$ is the radius.

You can re-arrange the above expression into

$$ x^2 + y^2 -2 a x - 2 b y = r^2 -a^2 - b^2 $$

which you can use to match the coefficients from the equation given.

$$ x^2 + y^2 + A x + B y = C$$

The coefficient $A$ in front of $x$ yields the coordinate $a$, as in $A x = -2 a x$ or $$a=-A/2$$

The coefficient $B$ in front of $y$ yields the coordinate $b$, as in $B y = -2 b y$ or $$b=-B/2$$

and finally the constant $C$ on the right hand side yields the radius $r$, as in $C = r^2 - a^2 - b^2$ or $$r = \sqrt{ C+a^2+b^2 }$$