Find the equation of the circle(s) touching $u$ and $S$.

Question

Given line $ u : 2x + y - 4 =0$ and circle $ S : x^2 + y^2 -16x -20y + 159 = 0$.
Find the equation of the circle touching $u$ and $S$ externally and has radius $2\sqrt5$.

What I have done is that I have considered the center to be $(a,b)$ and then found out the distance of it from the given line , which gives me to equations. I have also found the distance between the two centers (which is $3\sqrt5$).
On substituting , I found two values of $a$ and $b$ . Now how do I know which is the correct center without plotting on a graph.

Answer 1

Firstly you need to find all equations.

Let $(a,b)$ be a center of the circle.

Thus, since the equation of the given circle is $$(x-8)^2+(y-8)^2=5,$$ we get a following system: $$\frac{|2a+b-4|}{\sqrt5}=2\sqrt5$$ and $$(a-8)^2+(b-10)^2=45.$$ The first equation gives $b=14-2a$ or $b=-6-2a$.

Let $b=14-2a$. In the second case you make the same work.

Thus, $$(a-8)^2+(4-2a)^2=45,$$ which gives $a=5$ or $a=\frac{7}{5}$.

Let $a=5$. The case $a=\frac{7}{5}$ is similar.

Thus, $b=4$

Now, let $M$ be a touching point.

We need that $M$ is placed between $O_1(8,10)$ and $O_2(5,4)$ such that $O_1M:MO_2=1:2$.

Thus, $$M\left(\frac{8\cdot2+5}{3},\frac{2\cdot10+4}{3}\right)$$ or $$M(7,8)$$ and easy to check that $M$ placed on our circles: $$(7-5)^2+(8-4)^2=20$$ and $$(7-8)^2+(8-10)^2=5,$$ which says that $$(x-5)^2+(y-4)^2=20$$ is valid.

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