One of the two circles $x^2+y^2+\lambda_1(x-y)+c=0$ where $\lambda_1:\lambda_2\in\mathbb R$ and $x^2+y^2+\lambda_2(x-y)+c=0$ lies within the other

Question

If one of the two circles $x^2+y^2+\lambda_1(x-y)+c=0$ where $\lambda_1:\lambda_2\in\mathbb R$ and $x^2+y^2+\lambda_2(x-y)+c=0$ lies within the other then

• A) $c\lt0$
• B) $c=0$
• C) $c\gt0$
• D) none of these

My Attempt:

With $c=0$, circles will be touching each other at origin. So, maybe in technical sense, we may not say one is lying within another.

I am not able to come with any $c\lt0$ such that one circle is within another.

I am able to come up with the following case where $c\gt0$ and one circle is within another.

$x^2+y^2+4(x-y)+1=0$

$x^2+y^2+2(x-y)+1=0$

So, I think the answer should be option C).

What is the formal way of solving this question?

Answer 1

We can do the following factorization \begin{align} x^2+y^2+\lambda(x-y)+c&=0\\ \left(x^2+\lambda x+\frac{\lambda^2}{4}\right)+\left(y^2-\lambda y +\frac{\lambda^2}{4}\right)+\left(c-\frac{\lambda^2}{2} \right)&=0\\ \left(x+\frac{\lambda}{2}\right)^2+\left(y-\frac{\lambda}{2}\right)^2&=\frac{\lambda^2}{2}-c \end{align} so the first circle has center $(-\frac{\lambda_1}{2},\frac{\lambda_1}{2})$ and radius $\sqrt{\frac{\lambda^2_1}{2}-c}$, and the second has center $(-\frac{\lambda_2}{2},\frac{\lambda_2}{2})$ and radius $\sqrt{\frac{\lambda^2_2}{2}-c}$. So the distance between their centers is $\frac{1}{\sqrt{2}} |\lambda_1-\lambda_2|$. One will be inside the other if the largest radius is larger than the distance between them plus the smaller radius (we can see this geoemtrically). Lets assume without loss of generality that $\lambda_1\geq\lambda_2$, so we need \begin{align} \sqrt{\frac{\lambda^2_1}{2}-c}&>\frac{1}{\sqrt{2}} |\lambda_1-\lambda_2|+\sqrt{\frac{\lambda^2_2}{2}-c}\\ \frac{\lambda^2_1}{2}-c&>\frac{\lambda_1^2+\lambda_2^2-2\lambda_1\lambda_2}{2}+\frac{\lambda^2_2}{2}-c+(\lambda_1-\lambda_2)\sqrt{\lambda^2_2-2c}\\ \lambda_2(\lambda_1-\lambda_2)&>(\lambda_1-\lambda_2)\sqrt{\lambda^2_2-2c}\\ \lambda_2^2&>\lambda^2-2c\\ c&>0 \end{align} where when going from the third to second last line we took care to check that everything that needs to be positive was, enforced by our $\lambda_1\geq\lambda_2$ assumption.

Answer 2

Two circles are

• either intersecting in two points (including the limit case where these two points "coalesce"). In this case the line joining these two points (which is called the "radical axis" is known to be obtained by forming the difference of the two equations (or the tangent in the coalescing case).

• or non intersecting, which is equivalent to say that one circle is inside the other.

Here, let is form the difference equation ; we get

$$(\lambda_2-\lambda_1)(x-y)=0 \ \iff \ y=x \tag{1}$$

The intersection of this line with the two circles is obtained by replacing $y$ by $x$ in the two equations :

$$x^2+x^2+c=0\tag{2}$$

• In the case $c>0$, (2) has no solution : therefore, there is no intersection : the radical axis is external to both circles.

• In the case $c \le 0$, we get $x=y=\sqrt{-c/2}$ :

$$-c+\lambda_k\sqrt{-c/2}(x-x)+c=0$$

meaning that there are intersection points.