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Prove that $|z-10|=3|z-2$| is the equation of a circle with radius $3$ and center $1$.

I tried the following solution, with no result:

Considering $z=x+yi$,

$$|x+yi-10|=3|x+yi-2|$$

$$|(x-10)+yi|=3|(x-2)+yi|$$

$$\sqrt{(x-10)^2+y^2}=3\sqrt{(x-2)^2+y^2}$$

power of $2$:

$$(x-10)^2+y^2=9((x-2)^2+y^2)$$

after simplification, we will have:

$$8x^2+8y^2-16x-64=0$$

or: $$x^2+y^2-2x-8=0$$

I tried $z=x-yi$, as well. But there is no way I can come up with the intended result, that is, radius of $3$ and center of $1$. Any help would be appreciated.

Ted Shifrin
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Amethyst
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  • The locus of points in the plane with constant ratio $,\left|\dfrac{z-10}{z-2}\right| = 3,$ between the distances to the two fixed points $,10, 2,$ is a circle of Apollonius having as diameter the points that divide the segment between $,10,$ and $,2,$ in ratio $,3:1,$ internally and respectively externally. See this answer for the general proof. – dxiv Feb 21 '18 at 08:09

1 Answers1

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You are almost at end of your exercise. Just notice that $$0=x^2+y^2-2x-8=(x-1)^2+y^2-9=|z-1|^2-3^2\Leftrightarrow |z-1|=3$$ where $z=x+iy$.

Robert Z
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