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Further Maths, Loci URGENT HELP

hey
I have a question that I have been trying to solve it but I am getting unusual outcome,
Can a radius of a circle could possible be an imaginary? is it possible?
It's possible yes, but I'm not sure they'd give you an imaginary radius in a FM loci question, what was the question?
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MathsLove
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Original post by psycholeo
It's possible yes, but I'm not sure they'd give you an imaginary radius in a FM loci question, what was the question?


|z-6-i|=2|z-9-4|
Do you mean 2 l z -9 - 4i l ?
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MathsLove
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Original post by psycholeo
Do you mean 2 l z -9 - 4i l ?


Yes sorry
Original post by MathsLove
Yes sorry

The equation specified is not a circle, it's the equation of a perpendicular bisector in the form l z - a l = l z - b l where a and b are points
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MathsLove
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Original post by psycholeo
The equation specified is not a circle, it's the equation of a perpendicular bisector in the form l z - a l = l z - b l where a and b are points


You left out the number in front of the modules
That’s why a circle
Original post by MathsLove
|z-6-i|=2|z-9-4|


The radius here is 8\sqrt{8}, why are you asking about imaginary radii?

But to answer your question, it doesn't make sense to say a radius is an imaginary number. The radius is defined to be r≥0r\geq 0 where r∈Rr \in \mathbb{R} in order to have a circle. An imaginary radius would imply no solutions, as then you would have that r2<0r^2 < 0 which makes no sense as on the other side in eq x2+y2=r2x^2 + y^2 = r^2 you are adding squares of two real numbers and getting a -ve. You would need to let x,y∈Cx,y \in \mathbb{C} for the equation to make sense, but then you cannot define zz as x+iyx+\mathbf{i}y
(edited 5 years ago)
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MathsLove
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Original post by RDKGames
The radius here is 8\sqrt{8}, why are you asking about imaginary radii?

But to answer your question, it doesn't make sense to say a radius is an imaginary number. The radius is defined to be r≥0r\geq 0 where r∈Rr \in \mathbb{R} in order to have a circle. An imaginary radius would imply no solutions, as then you would have that r2<0r^2 < 0 which makes no sense as on the other side in eq x2+y2=r2x^2 + y^2 = r^2 you are adding squares of two real numbers and getting a -ve. You would need to let x,y∈Cx,y \in \mathbb{C} for the equation to make sense, but then you cannot define zz as x+iyx+\mathbf{i}y


Thank you very much

I think I inspected the outcome too early which putted me off, now when I worked to the bottom the radius is not an imaginary lol when I had a huge negative numbers in the RHS I had a panic attack saying “how am I gonna increase this negative numbers to positive “ lol

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