Consider the circles $x^2+y^2=1$ and $x^2+y^2-2x-6y+6=0$. Find the equations of common tangents to the two circles.

Question

Applying condition of tangency

$$y=mx\pm \sqrt {1+m^2}$$

$$y-3=m(x-1)\pm 2\sqrt{1+m^2}$$

So $$\pm \sqrt {1+m^2}=3-m\pm 2\sqrt{1+m^2}$$ $$3-m=\pm \sqrt{1+m^2}$$

$$m=\frac 43$$

So there should be one tangent. From the diagram, it is obvious that it will have 4 tangents, but why didn’t they show up here? (There are 4 answers to this question)

Edit- Checking back, I saw that it was a problem with the $\pm$. Adjusting for that, I got 2 more $m=0, -\frac 34$. But I am missing the line $x+1=0$

Answer 1

You should use instead the expressions below for the tangents of the two circles,

$$y\cos\theta=x\sin\theta \pm 1\tag 1$$ $$(y-3)\cos\theta=(x-1)\sin\theta \pm 2$$

where $\theta$ represents the tangent angle and is defined for all cases. Combine the equations for the four common tangents,

$$3\cos\theta-\sin\theta = \pm 1, \> \pm 3$$

Solve to obtain the four tangent angles,

$$\theta = 0, \> \frac\pi2,\> \tan^{-1} \frac43, \> -\tan^{-1} \frac34$$

Then, substitute the angles into the tangent equations (1) to obtain the four tangent lines,

$$y=1, \>\>\>\>\> x = -1, \>\>\>\>\> y = \frac43 x - \frac53, \>\>\>\>\>y = -\frac34 x + \frac54$$

Answer 2

Because $x=-1$ is also a common tangent.

Since this tangent has no a slope, you can not get it from the equation $y=mx+n$.

I like the following way.

Let $y=mx+n$ or $mx-y+n=0$ be an equation of the tangent.

Thus, $$\frac{|m\cdot0-0+n|}{\sqrt{m^2+1}}=1$$ and $$\frac{|m\cdot1-3+n|}{\sqrt{m^2+1}}=2$$ or $$|n|=\sqrt{m^2+1}$$ and $$|m+n-3|=2\sqrt{m^2+1},$$ which gives $$|m+n-3|=2|n|...$$

Answer 3

$S_1:x^2+y^2=1,~~~ S_2:x^2+y^2-2x-6y+6=0$

There are wto circles with centers $C_1(0,0), C_2(1,3)$ with radie aas $r_1=1, r_2=2$, respectively.

Two non-intersecting circles have four common tangents 2 direct and two transverse. Transverse tangents meet at point T internally dividing $C_1$ and $C_2$ in ratio $r_1:r_2$. Direct tangents meet at D which divides $C_1$ and $C_2$ externally in the same ratio.

Here T and D work out to be $(1/3,1)$ and $(-1,-3)$ respectively.

The family of lines passing through T are $y-1=m(x-1/3)~~~(1)$,for tangency perpendicular to it from $C_1$ should equal to $r_1$. Then we get $$\left |\frac{-1+m/3}{\sqrt{1+m^2}}\right|=1 \implies m=0,-3/4~~~(2)$$ Squaring and solving we get the two transverse tangents from (1) as $$y=1, ~~~3x+4y=5~~~(3)$$

Next family of lines passing through $D(-1.-3)$ is: $y+3=M(x+1)~~~(4)$ for tangency perpendicular to (3) from $(0,0)$ should be equal to $r_1$ again, then we get $$\left| \frac{3-M}{\sqrt{1+M^2}}\right|=1~~~(5) \implies M=\infty, M=4/3~~~(5)$$ $M=\infty$ satisfies in a limiting way. So from (4), we get direct tangents as $$x=-1,~~ 4x-3y-5=0~~~(6).$$ See the Fig.below: