Number of complex $z$ such that $|z+1|= |z+i|$ and $|z|=5$?

Question

How many complex numbers $z$ are there such that $|z+1|=|z+i|$ and $|z|=5$?

My attempt :

I got $2$, that is $ z=-2, z= +2$ , $|z| = {\sqrt{ 2^2+1}}$, $|z| = {\sqrt{(-2^2) +1}}$

Is it true ?

Answer 1

From $|z+1|= |z+i|$ we get $z$ at equaly distance from $-1$ and $-i$ so $z$ is on perpendicular bisetor for segment between $-1$ and $-i$, that is line $y=x$ so $$z=x+xi$$ for some real $x$. Then $|x|\sqrt{1+1} = 5$ so $x=\pm{5\sqrt{2}\over 2}$

Answer 2

Let $A(-1,0)$ and $B(0,-1)$.

Thus, we need to find a number of intersect points of the perpendicular bisector of $AB$

with the circle $x^2+y^2=25.$

Now we see that they are indeed two points because $A$ and $B$ are placed inside the circle.