# System of equations with complex numbers-circles

The system of equations \begin{align*} |z - 2 - 2i| &= \sqrt{23}, \\ |z - 8 - 5i| &= \sqrt{38} \end{align*} has two solutions $z_1$ and $z_2$ in complex numbers. Find $(z_1 + z_2)/2$.

So far I have gotten the two original equations to equations of circles, $(a-2)^2 +(b-2)^2=23$ and $(a-8)^2+(b-5)^2=38$. From here how do I find the solutions? Thanks.

• Why do you need the original circle equations? – Koba Jun 12 '15 at 21:42
• The point of this exercise is that you realize that $(z_1+z_2)/2$ must lie on the line that connects the centres of the two circles. – M. Wind Jun 12 '15 at 22:01
• Gotcha. I was writing an answer solving those by letting $z=a+bi$. – Koba Jun 12 '15 at 22:07

The first thing to do is to make a drawing of the two circles. Label their centres $C_1$ and $C_2$, and the two intersections of the circles $I_1$ and $I_2$. Now connect these four points by straight lines, and label as $P$ the point half-way between $I_1$ and $I_2$, which is on the line connecting $C_1$ and $C_2$.

We recognize that we have four triangles, all with a $90$ degree angle. So we can apply Pythagoras. For convenience we label the line-piece $C_1$ to $P$ as $a$; $C_2$ to $P$ as $b$; $I_1$ (or $I_2$) to $P$ as $c$. Now:

$$(a+b)^2 = 45$$

$$a^2 + c^2 = 23$$

$$b^2 + c^2 = 38$$

We eliminate $c^2$ by subtracting the second equation from the third, yielding:

$$(b-a)(b+a) = 15$$

Dividing the first equation by this result gives

$$\frac {b+a}{b-a} = 3$$

From which it follows that $b = 2a$. So $a$ is one-third of the distance between $C_1$ and $C_2$. Therefore the position of point $P$ is given by $2+2i$ + $(6+3i)/3$ = $4+3i$.

• Sorry if this should be clear, but I'm not really sure how we know we have four right triangles. – jjhh Jan 15 '18 at 0:01

You're given the distances of the solution points to $A = 2+2i$ and $B = 8+5i.$

One of the solutions and these two complex numbers give you a triangle in the complex plane, and you know the lengths of all the sides.

We can use Law of Cosines to find the angle $\theta$ that has $2+2i$ as its vertex:

$$38 = 45 + 23 - 2 \sqrt{45} \sqrt{23} \cos \theta \to \cos \theta \approx. 0.466252.$$

The projection of $AZ_1$ onto $AB$ is then $\sqrt{23} \cdot 0.466252 \approx 2.23606$.

Why calculate this? Because the other solution is symmetric on the other side of $AB$, and what you're asked to find, in essence, is the average of the two solutions, which lies on $AB$.

The unit vector from $A$ to $B$ is $(6 + 3i)/\sqrt{45} = (2+i)/\sqrt{5}.$

So, the solution is

$$(Z_1 + Z_2)/2 = 2 + 2i + 2.23606 \cdot (6 + 3i)/\sqrt{45} \approx 4 + 3i.$$

(The fact that this worked out so nicely makes me think there was an easier way, but oh well ...)

• nice solution. Though one of the tags is algebra precalculus, so I would think there must be an easier way. I think it definitely has to do with finding the midway point between the centers of two circles. – Koba Jun 12 '15 at 22:25

We can also treat this in a way that has little to do with complex numbers. The line connecting the centers of the circles at $\ (2, \ 2) \$ and $\ (8, \ 5 ) \$ has a slope of $\ \frac{5 - 2}{8 - 2} \ = \ \frac{1}{2} \$ and has the equation

$$y \ - \ 2 \ = \ \frac{1}{2} \ (x - 2) \ \ \Rightarrow \ \ y \ = \ \frac{1}{2} \ x \ + \ 1 \ \ .$$

Using a theorem from classical geometry, the two intersection points of the two circles lie on a mutual chord, of which the line connecting the centers of the circles is its perpendicular bisector (a fact also used in some way by M. Wind and John). Thus, we expect the chord to have a slope of $\ -2 \$ .

A relation for the coordinates of the intersection points of the circles can be found by subtracting one circle equation from the other; thus,

$$\ x^2 \ - \ 16x \ + \ 64 \ + \ y^2 \ - \ 10y \ + \ 25 \ = \ 38$$ $$- \ ( \ x^2 \ - \ 4x \ + \ 4 \ + \ y^2 \ - \ 4y \ + \ 4 \ = \ 23 \ )$$ $$--------------------$$ $$-12x \ + \ 60 \ - \ 6y \ + \ 21 \ = \ 15$$

$$\Rightarrow \ \ y \ = \ -2x \ + \ 11 \ \ .$$

The chord and its perpendicular bisector meet at

$$\frac{1}{2} \ x \ + \ 1 \ = \ -2x \ + \ 11 \ \ \Rightarrow \ \ \frac{5}{2} \ x \ = \ 10 \ \ \Rightarrow \ \ x \ = \ 4 \ \ \Rightarrow \ \ y \ = \ 3 \ \ ,$$

found by inserting the value for $\ x \$ into either of the linear equations we've determined. Hence, the midpoint of the chord containing the intersections of the circles lies at $\ z \ = \ 4 \ + \ 3i \$ in the complex plane.

Though we aren't asked to find them, we can also locate $\ z_1 \ \ \text{and} \ \ z_2 \$ . Inserting $\ y \ = \ -2x \ + \ 11 \$ into the first circle equation, we produce

$$(x - 2)^2 \ + \ (-2x + 11 - 2)^2 \ = \ 23 \ \ \Rightarrow \ \ 5x^2 \ - \ 40x \ + \ 62 \ = \ 0 \ \ ,$$

for which the roots are $\ x \ = \ 4 \ \pm \ \frac{3}{5} \ \sqrt{10} \$ . [Using the other circle equation instead yields exactly the same quadratic equation.]

We may then calculate $\ y \ = \ -2 \ (4 \ \pm \ \frac{3}{5} \ \sqrt{10}) \ + \ 11 \ = \ 3 \ \mp \ \frac{6}{5} \ \sqrt{10} \$ . Thus, we find

$$z_1 \ = \ (4 \ - \ \frac{3}{5} \ \sqrt{10}, \ 3 \ + \ \frac{6}{5} \ \sqrt{10}) \ \ , \ \ z_2 \ = \ (4 \ + \ \frac{3}{5} \ \sqrt{10}, \ 3 \ - \ \frac{6}{5} \ \sqrt{10}) \ \ .$$

(Some brief work with right triangles then tells us that the length of the chord between $\ z_1 \ \ \text{and} \ \ z_2 \$ is $\ \sqrt{\frac{36 + 144}{25} \ \cdot \ 10} \ = \ \sqrt{72} \ = \ 6 \ \sqrt{2} \$ .)