Solving for equation of a circle when tangents to it are given.

Question

Given: $\mathrm L(x) = x - 2$, $\mathrm M(y) = y - 5$ and $\mathrm N(x, y)= 3x - 4y - 10$.

To find: All the circles that are tangent to these three lines.

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Outline of the method :

If we parameterised any line $\mathrm Z$ in terms of $x(t) = pt + q$ and $y(t) = rt + s$, then for a general circle $\mathrm C(x, y) = (x - \alpha)^2 + (y - \beta)^2 - \gamma^2$ we can substitute the the parameters of the line and get

$$\mathrm C(x(t), y(t)) = a t^2 + bt + c = 0,$$ where $a,b,c \in \Bbb R$ and they depend on $p$,$q$.

Now if this quadratic has $2$ roots then the circle intersects the line at $2$ points and same for $1$ and $0$ roots.

If we are given that $\mathrm Z$ is tangent then we can using completing the square and get $(t - d(p, q) ) ^2 + e(p, q) = 0$.

So our condition for the line tangent to the circle would be $e(p, q) = 0$.

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Parameterising the three equations,

$\mathbf L^\prime (l) = (2,0) + l (0,1)$, $\mathbf M^\prime(m) = (0, 5) + m(-1, 0)$ and $\mathbf N^\prime(n) = (2 , -1) + n(4,3)$.

Substituting these in the general equation of a circle, $\mathrm C(x, y) = (x - h)^2 + (y - k)^2 - r^2$.

$$\left\{ \begin{align} \mathrm{C(L_x, L_y)} &= (2-h)^2 +(l-k)^2 -r^2 \\ \mathrm{C(M_x, M_y)} &= (m+h)^2 +(5-k)^2 -r^2 \\ \mathrm{C(N_x, N_y)} &= (2+4n-h)^2 +(3n-1-k)^2 -r^2 \end{align} \right.$$

As per the above method,

From first equation we get $r^2 = (2 - h) \iff r = |2 - h|$ and from second equation, $r = |5 - k|$.

Completing the square on the third equation we get,

$$\mathrm{C(N_x, N_y)} = \left(5n + \dfrac{5 - 4h - 3k}{5}\right) + 5 - 4h + 2k - r^2 + h^2 + k^2 - \left[1 - \dfrac{4h - 3k}{5}\right]^2.$$

Therefore,

$$ 2 + 2k + h^2 + k^2 = r^2 + 4h + \left[1 - \dfrac{4h - 3k}{5}\right]^2.$$

So here we got three equations:

\begin{align} 2 +2k +h^2 +k^2 &= r^2 +4h +\left[1 - \dfrac{4h-3k}{5}\right]^2 \\ r &= |5-k| \\ r &= |2-h|. \end{align}

I don't mind solving the third equation after substituting for $r$ from first in it, it is just a quadratic. What I don't understand is whether I should take $r = + (5 - k)$ or $r = - (5 - k)$ and $r = +(2 - h)$ or $r = - (2 - h)$?

Which two out of those should I take? And why?

Answer 1

You have by distance from a point to a line formula, $$|h-2|=|k-5|=\frac{|3h-4k-10|}{5}=r.$$ Consequently, $h=2\pm r$, and $k=5\pm r$. So we have four cases

• $h=2+r$, $k=5+r$ implies $|3(2+r)-4(5+r)-10|=|r+24|=5r \implies r=6$. So the equation of the circle is $$\color{red}{(x-8)^2+(y-11)^2=36}.$$
• $h=2+r$, $k=5-r$ implies $|3(2+r)-4(5-r)-10|=|7r-24|=5r \implies r=2,12$. So the equation of the circles are $$\color{red}{(x-4)^2+(y-3)^2=4},$$ and $$\color{red}{(x-14)^2+(y+7)^2=144}.$$
• $h=2-r$, $k=5+r$ implies $|3(2-r)-4(5+r)-10|=|7r+24|=5r$, which does not have any solution such that $r >0$.
• $h=2-r$, $k=5-r$ implies $|3(2-r)-4(5-r)-10|=|r-24|=5r \implies r=4$. So the equation of the circle is $$\color{red}{(x+2)^2+(y-1)^2=16}.$$

Answer 2

converting the equation in the Hessian Normalform we have $$\frac{|3x-4y-10|}{5}=R$$ where $$(x,y)$$ denotes the middle Point of our searched circle; and it must be $$x=2+R,y=5-R$$

Answer 3

Consider $\triangle ABC$, $A=(2,5),B=(2,-1),C=(10,5)$ and its incircle and three excircles,

\begin{align} a&=10,\quad b=8,\quad c=6,\quad \\ r&=2,\quad r_a=12,\quad r_b=6,\quad r_c=4 ,\\ O_i&=(4,3),\quad O_a=(14,-7),\quad O_b=(8,11),\quad O_c=(-2,1) . \end{align}