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How do l find the min & max radius of a circle on an argand diagram

So the question is modulus of z - 3-5 i=2r and l need to find a range for r,i checked the mark scheme but still don't get it. Btw the answer is 3 root 2 over 2 is less than r less than root 26 over 2

Reply 1

Original post by territorial-osci
So the question is modulus of z - 3-5 i=2r and l need to find a range for r,i checked the mark scheme but still don't get it. Btw the answer is 3 root 2 over 2 is less than r less than root 26 over 2

Not sure youve given the full question in the OP, but sketching the circle and thinking about how it relates to the centre usually gives the insights.

Reply 2

https://pmt.physicsandmathstutor.com/download/Maths/A-level/Further/Core-Pure/Edexcel/CP1/Topic-Qs/Set-A/Ch.2%20Argand%20Diagrams.pdf Its question 2 on here, there is more to the question but I didn’t think it was relevant to my question, that’s my bad

Reply 3

Original post by territorial-osci
https://pmt.physicsandmathstutor.com/download/Maths/A-level/Further/Core-Pure/Edexcel/CP1/Topic-Qs/Set-A/Ch.2%20Argand%20Diagrams.pdf Its question 2 on here, there is more to the question but I didn’t think it was relevant to my question, that’s my bad

For some reason, I didnt get notified that youd replied. But the extra info makes more sense. For the two constraints you have

a circle center on 3+5i and the radius is 2r.

a half line starting at 2 and arg=3pi/4

So you need two points of intersection
https://www.desmos.com/calculator/41mxrhzrem
So you need r to be large enough so that the circle touches the half line (lower value) but its max radius is when the circle hits the z=2 point as larger than that value, there will be only a single point of intersection with the half line.

These sort of questions may look hard, but really theyre just (gcse) line intersecting circles with the added restriction that its just a half line due to the arg definition. For this question, once you understand how the min/max radius is generated, you simply use pythagoras/sub the point into the circle equation.
(edited 9 months ago)

Reply 4

That acc makes sense now, thanks

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