# Circles in complex plane.

Find the real value of a for which there is at least one complex number satisfying $|z+4i|=\sqrt{a^2-12a+28}$ and $|z-4\sqrt{3}|\lt a$.

## My solutions:-

• Graphical solution:-

$|z+4i|=\sqrt{a^2-12a+28}$ represents a circle with center at $A\equiv(0,-4)$ and radius $r_1=\sqrt{a^2-12a+28}$ and similarly, $|z-4\sqrt{3}|\lt a$ represents all the points inside the circle with center at $B\equiv(4\sqrt{3},0)$ and radius $r_2=a$. So, the distance between the centers of the circle is $\sqrt{(4\sqrt{3})^2+(4^2)}=8$

Now consider the graphs of $\sqrt{a^2-12a+28}$ and $a$. The graph of $\sqrt{a^2-12a+28}$ which is a hyperbola with its centre at $(6,0)$ and vertexes being $(6\pm 2\sqrt2,0)$

It is clear from the plot that $a\in (0,6-2\sqrt2]\cup [6+2\sqrt2,\infty)$ for $\sqrt{a^2-12a+28}\ge0$ and $a\gt 0$

Now, lets do some case study.

Case 1:- When $a \ge 6+2\sqrt2$

In this case the radius of the circle represented by $|z-4\sqrt3|=a$ is greater than the distance between the centers of the circles, i.e $r_2\gt AB$ as $r_2\gt 8$. So, the circle $|z-4\sqrt3|=a$ either encloses the circle $|z+4i|=\sqrt{a^2-12a+28}$(or the point $-4i$, which is the case when $a=6+2\sqrt2$) fully or encloses a portion of it and in both the cases we find that we obtain a number common to both the regions. So, $a\in(6+2\sqrt2,\infty)$

Case 2:- $0\le a\le 6-2\sqrt2$

In this case we see that due to the bounding of $a$, we see that $(r_1+r_2)\lt 8$, so there is no intersection of the wanted regions, so there is no solution in this region.

So the required interval of $a$ is $\boxed{a \in [6+2\sqrt2, \infty)}$

• Algebraic Solution(or whatever you wanna call it):-

Lets consider all the possible circles that can be drawn for the given circles, they are as given in the figure:-

From the above drawn circles we can consider the following cases.

Case 1:- From Figure-3 (I have not numbered it so consider it in the left to right manner) we can get the following condition $$AB\lt r_1+r_2 \implies 8 \lt \sqrt{a^2-12a+28} + a \implies a\gt 9$$ This case deals with the limiting condition of both the circles touching.

Case 2:- Now to not let the Figure-5 take place we have to consider the case $$AB\gt r_1-r_2 \implies 8 \gt \sqrt{a^2-12a+28} -a \implies a\gt -\frac{9}{7}$$

Now, I don't know what to conclude from this, so please do tell me that

Case 3:- Consider Figure-4(don't know why I drew the last figure as it IMO represents the same case as Figure 4), the case we get is $$AB\lt r_2-r_1 \implies 8\lt a-\sqrt{a^2-12a+28} \implies a\gt 9$$

## My deal with the question:-

$a\gt 9$, $a\gt 9$ everywhere not a single $a\ge 6+2\sqrt2$ to see.(It's a joke)

So as you can see that from the Algebraic solution I got $a\gt 9$ and from the Graphical approach to the question I get $a\ge 6+2\sqrt2$. And the book I am solving also gives the answer interval as $a\in (9,\infty)$. So what is wrong with the solutions and do point out the errors. As, always more elegant solutions are welcome.

In my opinion the answer should be $a\in [6+2\sqrt2, \infty)$, which is evident from the graph figure below.

The answer should be $a\ge 6+2\sqrt 2$.

($a\gt 9$ is not correct since, for $a=9$, $z=-3i$ satisfies the conditions.)

In your Algebraic Solution, it seems that you have some errors.

Case 1:- $$8 \lt \sqrt{a^2-12a+28} + a \implies a\gt 9$$ Case 2:- $$8 \gt \sqrt{a^2-12a+28} -a \implies a\gt -\frac{9}{7}$$ Case 3:- $$8\lt a-\sqrt{a^2-12a+28} \implies a\gt 9$$

These should be incorrect. Under $0\lt a\le 6-2\sqrt 2$ or $a\ge 6+2\sqrt 2$, I got the followings : $$8 \lt \sqrt{a^2-12a+28} + a \iff a\ge 6+2\sqrt 2$$ (To solve $8-a\lt \sqrt{a^2-12a+28}$, separate it into two cases : If $8-a\lt 0$, then the inequality holds(the LHS is negative, the RHS is non-negative). If $8-a\ge 0$, then since the both sides are non-negative, squaring gives $a\gt 9$. There is no $a$ such that $a\le 8$ and $a\gt 9$. Therefore, $a\gt 8$. Considering $0\lt a\le 6-2\sqrt 2$ or $a\ge 6+2\sqrt 2$, the solution is $a\ge 6+2\sqrt 2$.)

$$8 \gt \sqrt{a^2-12a+28} -a \iff 0\lt a\le 6-2\sqrt 2\quad\text{or}\quad a\ge 6+2\sqrt 2$$ $$8\lt a-\sqrt{a^2-12a+28}\iff 6+2\sqrt 2\le a\lt 9$$

In the following, I will write a solution in the similar way as your Algebraic Solution.

We consider the two circles $$C_1 : x^2+(y+4)^2=a^2-12a+28\qquad\text{and}\qquad C_2 : (x-4\sqrt 3)^2+y^2=a^2$$ where $0\lt a\le 6-2\sqrt 2$ or $a\ge 6+2\sqrt 2$.

The distance between the centers is $8$.

Case 1 : The two cirlces have no intersection points, and no circle is inside the other : $8\gt \sqrt{a^2-12a+28}+a\iff 0\lt a\le 6-2\sqrt 2$. This case has to be eliminated.

Case 2 : The two circles have the only one intersection point, and no circle is inside the other : $8=\sqrt{a^2-12a+28}+a$. There is no such $a$.

Case 3 : The two circles have two intersection points : $|\sqrt{a^2-12a+28}-a|\lt 8\lt \sqrt{a^2-12a+28}+a\iff a\gt 9$. This case is sufficient.

Case 4 : The two circles have the only one intersection point, and one circle is inside the other : $8=|\sqrt{a^2-12a+28}-a|$.

Case 4-1 : The case when $(4\sqrt 3,0)$ is inside $C_1$. There is no such $a$.

Case 4-2 : The case when $(0,-4)$ is inside $C_2$ : $a=9$. This case is sufficient.

Case 5 : The two circles have no intersection points, and one circle is inside the other : $8\lt |\sqrt{a^2-12a+28}-a|$

Case 5-1 : The case when $(4\sqrt 3,0)$ is inside $C_1$ : There is no such $a$.

Case 5-2 : The case when $(0,-4)$ is inside $C_2$ : $8\lt a\lt 9$. This case is sufficient.

Therefore, the answer is $a\ge 6+2\sqrt 2$.

• From the inequality in Case 1, I am still getting $a>1$. I solved the inequality as follows:-$$8\gt\sqrt{a^2-12a+28}+a \implies 8-a\gt\sqrt{a^2-12a+28}\implies 64+a^2-16a\gt a^2-12a+28\implies 36\gt 4a \implies a\lt 9$$ Jul 23, 2016 at 6:57
• @user350331: See what I added in the answer. Ah, you mean case 1 in my solution, right? I added the details for the case 1 in your solution. Anyway, I hope that helps. Jul 23, 2016 at 7:06
• @user350331: For the case 1 in my solution, when you have $8-a\gt \sqrt{a^2-12a+28}$, you have to have $8-a\gt 0$ because of $\sqrt{a^2-12a+28}\ge 0$. Jul 23, 2016 at 7:18
• @user350331: OK, I'll tell you something if I find a good one. Just a tip : in each step, see if each side is positive or negative. Also, note that $\sqrt{\text{something}}$ gives that $\sqrt{\text{something}}\ge 0$, and $\text{something}\ge 0$. Jul 23, 2016 at 7:28
• @user350331: You can find tags (book-recommendation) (reference-request). You may want to see the questions with these tags to know what good questions (well-received) with these tags are. Jul 23, 2016 at 15:35