Understanding how the equation $(x-6)^2+(y-9)^2+ \lambda (3x-y-9)=0$ allows us to find the circle through $(0,1)$ tangent to $3x-y-9=0$ at $(6,9)$

Question

I'm required to find the equation of the circle passing through (0, 1) and touching the line $3x-y-9=0$ at $(6,9)$. Apparently, this is easily done by writing it out as a second degree equation; $$(x-6)^2+(y-9)^2+ \lambda (3x-y-9)=0$$ and solving it using the point.

I don't understand how this works; yes, it's a second-degree equation that sort of looks like a circle that we eventually get, but I can't grasp the rigor behind it. The equation seems meaningless; it specifies the set of points for which the square of their distances from (6, 9) equals their algebraic distances from the line multiplied by some constant.

How exactly do you get to that equation?

A related question has been asked on here before, but the method was only briefly touched upon, and incorrectly used, at that; thus the new post.

Answer 1

The equation [...] specifies the set of points for which the square of their distances from $(6,9)$ equals their algebraic distances from the line multiplied by some constant.

Correct! (What's incorrect is characterizing this relation as "meaningless".) So, you've basically answered your own question, which is a big time-saver. :)

The key is understanding the nature of that constant.

Consider a circle of radius $r$ tangent to a line at $T$ (with $T'$ the diametrically-opposite point), and a point $P$ on the circle whose projection onto the line is $Q$.

Defining $p:=|PT|$ and $q:=|PQ|$, we have, because $\triangle PQT\sim\triangle TPT'$, $$\begin{align} \frac{q}{p}=\frac{p}{2r} &\quad\to\quad p^2=q\cdot 2r \tag{1}\\ &\quad\to\quad (\text{dist to $T$})^2 = (\text{dist to line})\cdot(\text{constant}) \tag{2} \end{align}$$

So, the situation is (quite meaningfully) exactly as you describe. $\square$

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Importantly, the $\lambda$ in
$$(x-6)^2+(y-9)^2+\lambda(3x-y-9) = 0 \tag{3}$$ doesn't correspond to $2r$, because $3x-y-9$ doesn't quite correspond to the distance to the line.

• For one thing, $3x-y-9$ is a signed value (positive for points on one side of the line, negative for points on the other side). So, $\lambda$ is incorporating the sign needed accommodate for this.

• For another, the distance-to-line formula requires dividing-out by a normalizing factor (the square root of the squares of the coefficients of $x$ and $y$; here, $\sqrt{3^2+1^2}=\sqrt{10}$). So, $\lambda$ is also incorporating this value.

A more-descriptive form of $(3)$ would be $$(x-6)^2+(y-9)^2 = \pm\frac{2r}{\sqrt{10}}(3x-y-9) \tag{4}$$ We choose $r$ to specify the size of the circle, and sign to specify on which side of the line it lies.

In the context of the problem, being given that the circle passes through $(0,1)$ forces the size and sign choices. (Of course, in the context of the problem, if "all we want" is the equation, then we can just use the less-descriptive form, $(3)$.)

Answer 2

$(x-6)^2+(y-9)^2+ \lambda (3x-y-9)=0$ is a equation of a circle.

Suppose this circle touches the line $3x-y-9=0$ at a point $(a,b)$ other than $(6,9)$ then $3a-b-9=0$ but $(a-6)^2+(b-9)^2 \ne 0$.
This contradicts $(a-6)^2+(b-9)^2+ \lambda (3a-b-9)=0$.

Hence the circle $(x-6)^2+(y-9)^2+ \lambda (3x-y-9)=0~$ touches $~3x-y-9=0$ only at $(6,9)$, $~$so $~3x-y-9=0~$ is a tangent to the circle.

Answer 3

Suppose that you have a line $r:ax+by+c=0$ and you know that a circle $C$ is tangent to $r$ at the point $(p,q)$.

A property of circles is the tangent line $s$ through a point $P$ is perpendicular to the line $\overline{OP}$ (where $O$ is the center of the circle). The line $r'$ perpendicular to $r:ax+by+c=0$ passing through $P=(p,q)$ has equation $$b(x-p)-a(y-q)=0$$ and a general point of $r'$ can be written as $Q=(p+\lambda a,q+\lambda b)$.

So, the a general circle tangent to $r$ at $(p,q)$ has equation (where $(x_0,y_0)=(p+\lambda a,q+\lambda b)$ is the center) $$(x-x_0)^2+(y-y_0)^2-R^2=0$$with $x_0=p+\lambda a,y_0=q+\lambda b$ and $R=|OP|=|\lambda|\sqrt{a^2+b^2}$, so the equation becomes $$\underbrace{(x-p)^2-2(x-p)\lambda a+{(\lambda a)^2}}_{=(x-p-\lambda a)^2}+\underbrace{(y-q)^2-2(x-q)\lambda b+(\lambda b)^2}_{(x-q-\lambda b)^2}-{\lambda^2(a^2+b^2)}=0$$ now you can cancel $(\lambda a)^2,(\lambda b)^2$ and $\lambda^2(a^2+b^2)$, so the equation of this family of cicles became $$(x-p)^2+(y-b)^2-2\lambda[a(x-p)+b(y-q)]=0$$setting $\lambda'=-2\lambda$ and using that $ap+bq=-c$, you reach your equation $$(x-p)^2+(y-q)^2+\lambda'[ax+by+c]=0$$