If the line $ax+by +c = 0$ touches the circle $x^2+y^2 -2x=\frac{3}{5}$ and is normal to $x^2+y^2+2x-4y+1=0$, what is (a,b)?

Question

I have a question that goes:

If the line $ax+by +c = 0$ touches the circle $x^2+y^2 -2x=\frac{3}{5}$ and is normal to $x^2+y^2+2x-4y+1=0$, what is (a,b)?

So what I tried was I know that since the line is normal to the 2nd circle, so it must pass through the center of the second circle which is $(-1,2)$.

So from that I got that $$-a+2b+c=0$$

But I cant really find any other equations here that would help, I tried differentiation the curves but I dont have the point of contact so can't really do anything there.

I also know that the tangent to the circle $x^2+y^2+2gx+2fy+c = 0$ at $(a,b)$ is $ax+by+(a+x)g+(b+y)f +c = 0$

I don't know how to proceed, can someone help?

Answer 1

By sketching a graph in mind, we see that the line $ax+by+c=0$ is just a line that is passing through $(-1,2)$ (the center of the circle $x^2+y^2+2x-4y+1=0$) and is tangent to the circle $x^2+y^2-2x=3/5$.

Therefore we looking for tangent lines from point $(-1,2)$ to the circle $x^2+y^2-2x=3/5$.

The line passing through $(-1,2)$ with slop $m$ is $$y=m(x+1)+2.$$ The intersects points of the line and the circle $x^2+y^2-2x=3/5$ is found by $$x^2+(m(x+1)+2)^2-2x=3/5$$ simplifying: $$(1+m^2)x^2+[2m(m+2)-2]x+(m+2)^2-3/5=0\tag{*}$$ Since we need the line $y=m(x+1)+2$ to touch the circle, then the discriminant of $(*)$ must be zero: $$(2m(m+2)-2)^2-4(1+m^2)((m+2)^2-3/5)=0$$ which simplifies to $$\frac{-4}{5}(m+3)(m+\frac{1}{3})=0$$ i.e. $m=-3$ or $m=-1/3$. Thus the tangent lines are $$y=-3(x+1)+2\quad\text{and}\quad y=-\frac13(x+1)+2$$

Answer 2

Another way.

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

Thus, $$mx-y+n=0,$$ which since equations of our circles they are $$(x+1)^2+(y-2)^2=4$$ and $$(x-1)^2+y^2=\frac{8}{5},$$ we obtain that the point $(-1,2)$ is placed on the line $y=mx+n$

and the distance from $(1,0)$ to the line is equal to $\sqrt{\frac{8}{5}}.$

Thus, $$-m-2+n=0$$ and $$\frac{|m\cdot1-1\cdot0+n|}{\sqrt{m^2+1}}=\sqrt{\frac{8}{5}},$$ which gives $$\frac{|2m+2|}{\sqrt{m^2+1}}=\sqrt{\frac{8}{5}}$$ or $$5(m+1)^2=2(m^2+1).$$ Can you end it now?

Answer 3

Let $(x_0,y_0)$ be the point of contact to $(x-1)^2+y^2=\frac85$. The tangent line equation is: $$y=y_0+y'(x_0)(x-x_0) \Rightarrow \\ y=y_0+\frac{1-x_0}{y_0}(x-x_0) \Rightarrow \\ \frac{1-x_0}{y_0}x-y+y_0-\frac{1-x_0}{y_0}x_0=0 \Rightarrow \\ a=\frac{1-x_0}{y_0};b=-1;c=y_0-\frac{1-x_0}{y_0}x_0$$ The tangent line passes through the point $(-1,2)$ (the center of the circle $(x+1)^2+(y-2)^2=4$).

So we make up the system: $$\begin{cases}2=y_0+\frac{1-x_0}{y_0}(-1-x_0)\\ (x_0-1)^2+y_0^2=\frac85\end{cases}\Rightarrow (x_0,y_0)=(-\frac15,-\frac25),(\frac75,\frac65).$$ Hence: $$(a,b,c)=(-3,-1,-1); (-\frac13,-1,\frac53).$$