Tangent to $x^2+y^2-6x-6y=-13$ and $x^2+y^2+2x+2y=-1$

Question

Considering the circles $\lambda: x^2+y^2-6x-6y=-13$ and $\theta: x^2+y^2+2x+2y=-1$ find the line simultaneously tangent to them.

I found the implicit derivative of those two, $\lambda: y'=-\frac{x-3}{y-3}$ and $\theta: y'=-\frac{x+1}{y+1}$ but then I don't know how to proceed.

I thought that setting $y'_{\lambda}=y'_{\theta}$ would solve, but it didn't.

If anyone could help I'll appreciate.

Answer 1

To prove that the point of intersection of the two direct common tangents lies on the line joining the centers of the two circles, notice that angles $\angle ACD$, $\angle AFE$, $\angle BDG$ and $\angle BEG$ are all right angles. Thus, $\angle CAB$+ $\angle DBA$ = $\angle FAB$ + $\angle ABE$ = $180^{\circ}$.
Also, $\angle GBD$ + $\angle DBG$ = $\angle GBD$ + $\angle CAB$ = $90^{\circ}$. Now conclude that $\angle DBG$ + $\angle DBA$ = $180^{\circ}$. Done.

Now, notice that by similarity of $\Delta CAG$ and $\Delta DBG$, $$\frac{AG}{BG}=\frac{AC}{BD}=\frac{R}{r}$$
By the external section formula, we have $$G\equiv \frac{R(-1,-1)-r(3,3)}{R-r}=(\frac{-R-3r}{R-r}, \frac{-R-3r}{R-r})$$ Now, the line may be written as $\displaystyle y+\frac{R+3r}{R-r}=m(x +\frac{R+3r}{R-r})$.
The distance of a point $(h,k)$ from a line $ax+by+c=0$ is $\left\vert \dfrac{ah+bk+c}{a^2+b^2}\right\vert$.
The distance from any center to the tangent is equal to the radius of that circle so write:$$\left\vert\dfrac{-1-m(-1)+(1-m) \frac{R+3r}{R-r}}{\sqrt{1+m^2}}\right\vert= 1$$ Now you will get two values of m and consequently, two common tangents. Oh, I forgot to add: R = $\sqrt 5$ and r=1.

Now I hope you will be able to find the transverse common tangents by yourself.

Answer 2

You are waving the fact that the two circumferences are different loci:
the tangent point on one will not have the same $(x,y)$ as on the other.

If you want to proceed on that way you shall write $$ \begin{array}{l} \left\{ \begin{array}{l} z_1 = - 13 = f_1 \left( {x_1 ,y_1 } \right) \\ z_1 = - 1 = f_2 \left( {x_2 ,y_2 } \right) \\ dz_1 = 0 = \frac{\partial }{{\partial x_1 }}f_1 \left( {x_1 ,y_1 } \right)dx_1 + \frac{\partial }{{\partial y_1 }}f_1 \left( {x_1 ,y_1 } \right)dy_1 \\ dz_2 = 0 = \frac{\partial }{{\partial x_2 }}f_2 \left( {x_2 ,y_2 } \right)dx_2 + \frac{\partial }{{\partial y_2 }}f_2 \left( {x_2 ,y_2 } \right)dy_2 \\ \end{array} \right.\quad \Rightarrow \\ \quad \Rightarrow \left\{ \begin{array}{l} z_1 = - 13 = f_1 \left( {x_1 ,y_1 } \right) \\ z_1 = - 1 = f_2 \left( {x_2 ,y_2 } \right) \\ \frac{{dy_1 }}{{dx_1 }} = - \frac{{\frac{\partial }{{\partial x_1 }}f_1 \left( {x_1 ,y_1 } \right)dx_1 }} {{\frac{\partial }{{\partial y_1 }}f_1 \left( {x_1 ,y_1 } \right)}} \\ \frac{{dy_2 }}{{dx_2 }} = \cdots \\ \end{array} \right. \\ \end{array} $$

Answer 3

From the givens, the first circle is

$(x - 3)^2 + ( y - 3) ^2 = 5 $

which is a circle centered at $(3,3)$ with a radius of $\sqrt{5}$

The second circle is

$ (x+1)^2 + (y+1)^2 = 1 $

a circle with center $(-1,-1)$ and radius $1$.

There are four possible lines of common tangency, external and internal. Let the corresponding points of tangency on the two circles be

$p = (3 + \sqrt{5} \cos(\theta) , 3 + \sqrt{5} \sin(\theta) ) $

and

$q = (-1 + \cos(\phi) , -1 + \sin(\phi) )$

Clearly, from the direction of the normal vector to the two circles, we have two cases: either $\theta = \phi$, or $ \theta = \phi + \pi $

In either case we want the vector $p - q$ to be perpendicular to the normal to the circle, thus we want,

$\big( 4 + \sqrt{5} \cos(\theta) - \cos(\phi) , 4 + \sqrt{5} \sin(\theta) - \sin(\phi) \big) \cdot ( \cos(\theta), \sin(\theta) ) = 0 $

In the the external tangents case, $\theta = \phi$, hence, the above dot product equation becomes

$ 4 \cos(\theta) + 4 \sin(\theta) + (\sqrt{5}-1) = 0 $

And this has two solutions. And they are

$ \theta = \dfrac{\pi}{4} \pm \cos^{-1}\left( \dfrac{1 - \sqrt{5} }{ 4 \sqrt{2} } \right) = 0.785398 \pm 1.79108 = -1.0057, 2.5765 $

With these two values we can compute the normal vector to the tangent lines as, respectively,

$n_1 = \langle \cos(-1.0057), \sin(-1.0057) \rangle = \langle 0.5355, -0.8445 \rangle$

$n_2 = \langle \cos(2.5765), \sin(2.5765) \rangle = \langle -0.8445, 0.5355 \rangle $

For the first of these two, a point on the second circle is

$q = (-1 + 0.5355, -1 - 0.8445 ) = (-0.4645, -1.8445)$

So the equation of the first (external tangent) is

$\boxed{0.5355 (x + 0.4645) - 0.8445( y + 1.8445) = 0}$

Similarly can derive the equation for the second of these two external tangent from a point on the second circle (or the first circle for that matter),

$q = (-1 -0.8445 , -1 + 0.5355 ) = (-1.8445, -0.4645)$

And the second equation is

$ \boxed{-0.8445 ( x + 1.8445) + 0.5355 (y + 0.4645 ) = 0} $

Next, we consider the internal tangents. Here $\theta = \phi + \pi$, therefore, $\cos(\phi) = -\cos(\theta), \sin(\phi) = - \sin(theta) $

so now the equation for orthogonality becomes

$ 4 \cos(\theta) + 4 \sin(\theta) + (\sqrt{5}+1) = 0 $

And it solution (details omitted) is

$ \theta_3 = -1.3944 $

$ \theta_4 = 2.9652 $

Now, the internal tangents normals, and points on them are

$ n_3 = (0.1755, -0.9845 )$

$ n_4 = (-0.9845, 0.1755)$

$p_3 = ( -1.1755 , -0.0155 )$

$ p_4 = (-0.0155, -1.1755 )$

So that the equations are

$\boxed{ 0.1755 (x + 1.1755) - 0.9845 (y + 0.0155) = 0 } $

and

$\boxed{ -0.9845 (x + 0.0155 ) + 0.1755 (y + 1.1755) = 0} $

The two circles and the tangents are shown in the figure below.

Answer 4

The dual conic method; a method that works for finding the common tangents of any two smooth conics.

$x^2+y^2-6x-6y=-13$ has dual $4X^2+18XY+6Y+4X^2+6X+1=0$

$x^2+y^2+2x+2y=-1$ has dual $2XY-2Y-2X+1=0$

The intersection points in the dual plane from i.e. the grobner basis $\langle 4Y^4+16Y^3-24Y^2+5, X+2Y^3+10Y^2-2Y-3\rangle$ are $P_1: (X,Y)=(-0.4090941234707588,0.6451620997691458),\\P_2: (X,Y)=(0.6451620997691458,-0.4090941234707588),\\P_3: (X,Y)=(0.9187629817678283,-5.154831199068685),\\P_4: (X,Y)=(-5.154831199068685,0.9187629817678283)$

corresponding to the common tangents $X(P_i)x+Y(P_i)y+1=0, i=1\ldots 4$ in the usual plane.