If the circles $x^2+y^2=c^2$ and $x^2+y^2+2ax=0$ touch each other, then prove that $4a^2=c^2$

Question

Radii of both the circles are $c$ and $a$ units.

Distances between the centre is $\sqrt{(0-a)^2+(0-0)^2}=a$

Since they touch each other $$a+c=a$$ $$c=0$$ what have I done wrong?

Answer 1

let $S_1 \equiv x^2+y^2-c^2 \equiv0$ and $S_2 \equiv x^2+y^2+2ax \equiv0$.

Now suppose $S_1,S_2$ touch externally. then,

$\sqrt {0^2+0^2+c^2}+\sqrt {0^2+a^2-0}=\sqrt {(0-0)^2+(0-a)^2} \implies c=0$ a contradiction.

Hence $S_1,S_2$ touch internally.

therefore, $|\sqrt {0^2+0^2+c^2}-\sqrt {0^2+a^2-0}|=|a|\implies2|a|=|c|\implies4a^2=c^2$

Answer 2

$(x+a)^2+y^2=a^2$ touches to $x^2+y^2=c^2,$ which says $$2|a|=|c|.$$

Draw it!

Answer 3

Without loss of generality we assume that $a$ and $c$ are positive.

The correct version should be $$4a^2=c^2$$

Note that the circle $$x^2+y^2+2ax=0$$ which is $$(x+a)^2+y^2=a^2$$ passes through the origin which is the center of the circle $$x^2+y^2=c^2$$.

In order for the circles to touch, we need to have $2a=c$, that is $$4a^2=c^2$$

Answer 4

We need $$||c|\pm|a||=|a|$$

$$\implies a^2=a^2+c^2\pm2|ca|$$

$$c^2=\mp|2ca|$$

Use $c\ne0,$

Answer 5

You assumed that the circles touch each other if and only if the distance between the centers is the sum of the radii of the circles. But that's just one of two possibilities; its the case in which the circles are boundaries of disks with a single common point. However, the van also be boundries of disks in which one of them is completely contained in the other one. And, in that case, what happens is that the distance between the centers is the difference of the radii of the circles.