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Find the Cartesian equation of the locus described by $|z+2-7i| = 2|z-10+2i|$ Write your answer in the form $(x+a)^2+(y+b)^2=k$.

This was a question from my end of year exams just gone and I'm unsure as to where I have gone wrong :(. If anyone could point me in the right direction I'd very much appreciate that.

Here was my working:

Let $z = x+iy$
$|z|$ = $\sqrt{x^2+y^2}$

$$|z+2-7i| = 2|z-10+2i| \\ \implies |(x+iy)+2-7i| = 2|(x+iy)-10+2i| \\ \implies |(x+2)(y-7y)i| = 2|(x-10)+(y+2)i| \\ \implies \sqrt{(x+2)^2+(y-7)^2i^2} = 2\sqrt{(x-10)^2+(y+2)^2i^2} \\ \implies (x+2)^2-((y-7^2) = 2((x-10)^2-(y+2)^2)$$

$$(x^2+4x+4)-(y^2-14y+49) = 2((x^2-20x+100)-(y^2+2y+4)) \\ \implies x^2+4x+4-y^2+14y-49 = 2x^2-40x+200-2y^2-4y-8 \\ \implies -237 = x^2-44x-y^2-18y\\ \implies x^2-44x-y^2-18 = -237$$

UmbQbify
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user535239
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3 Answers3

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Consider $|z+a+bi|=2|z+c+di|$. Squaring gives $$(x+a)^2+(y+b)^2=4(x+c)^2+4(y+d)^2$$ that is $$3(x^2+y^2)+(8c-2a)x+(8d-2b)y+4c^2+4d^2-a^2-b^2=0.$$ In general this is a circle. Now put in your values of $a,\ldots,d$.

Angina Seng
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HINT

Note that in general the equation

$$|z+a+bi|=k|z+c+di|\iff\frac{|z+a+bi|}{|z+c+di|}=\frac{d_1}{d_2}=k$$

describes a circle known as Circle of Apollonius.

Thus as an alternative you could find C and D by the given condition and then the center $O$ and radius $R$ of the circle

$$O=\frac{C+D}{2} \quad R=OD=OC$$

enter image description here

user
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  • As fully proved at the link posted in my comment under the original question. You should elaborate some more in order to make this into a proper answer. – dxiv Feb 26 '18 at 07:14
  • @dxiv my intent is to give a general interpretation of the given equation, for the solution of this particular case the OP receveid already many good hints and ways to solve, Thanks. – user Feb 26 '18 at 07:18
  • Sorry, but it's not an answer as written now, maybe a comment at best. – dxiv Feb 26 '18 at 07:19
  • @dxiv it is more an hint ang gives an idea to solve the problem by a different approach, finding C and D and the the center for example, is up to the OP decide if try this way or not – user Feb 26 '18 at 07:22
  • Your latest edit is a step forward towards making it into an answer. – dxiv Feb 26 '18 at 07:28
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    Gimusi.Your contribution to this problem is putting it in a wider context , Appolonius, and illustrating it very nicely. That is fine with me:))) – Peter Szilas Feb 26 '18 at 09:58
  • @PeterSzilas Thanks a lot for the appreciation Peter. Yes I think that this is a must have reference for this kind of problems. – user Feb 26 '18 at 10:27
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Consider z= (x+iy),Now put the value of Z in the given question. after applying rules of modulus we get `x^2+Y^2-28x+10y+121=0 Now we can arrange in square form like (x-14)^2+(y+5)^2=45

Shyam
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