Complex Locus Fun (1 Viewer)

[haboozin]

I think this question was pretty tricky...



Determine the values of K for which the simultaneous euqations:

|Z - (3 + 2i)| = 2 and |z - 2i| = k

have exactly one solution for z.


dont need to do working, i just thought it was tricky... see what you get.

 

[jake2.0]

do u use a graph?

 

[haboozin]

do u use a graph?

yea anyway u want..

 

[jake2.0]

k=1????????????

 

[Dumsum]

I'm thinking k=1 as well. And also k=5

 

[tez0r]

yeah, k=1 and k=5

 

[haboozin]

yea i missed the k= 5 when i did it :'(

 

[.ben]

how did you get the answers? simultaneous eqns?

 

[jake2.0]

i used a diagram, probably why i missed out on the k=5 bit

 

[Stefano]

What the?

I think this question was pretty tricky...



Determine the values of K for which the simultaneous euqations:

|Z - (3 + 2i)| = 2 and |z - 2i| = k

have exactly one solution for z.


dont need to do working, i just thought it was tricky... see what you get.

I got an infinite amount of solutions! :S

I'm imagining 2 circles:
1. Center: (3,2) & Radius: 2
2. Center: (0,2) & Radius: k

k could be anything though.... so couldn't circle 2 meet circle 1 at infinite places?

 

[jake2.0]

I got an infinite amount of solutions! :S

I'm imagining 2 circles:
1. Center: (3,2) & Radius: 2
2. Center: (0,2) & Radius: k

k could be anything though.... so couldn't circle 2 meet circle 1 at infinite places?

but you only want the two cirlces touching once

 

[Stefano]

but you only want the two cirlces touching once

Ah yes. I missed that part of the question. Thanks.

 

[Stefano]

Circle1: (x-3)^2 + (y+2)^2 = 2 --- &
Circle2: x^2 = (y-2)^2 = k --- @

Add 2 to b.s. of @ and make it equal to 2.

now, solve simultaneously...

after solving simultaneously i get:

k= 6x - 4y + 5

Is that correct thus far?
Where do we go next?

 

[haboozin]

Circle1: (x-3)^2 + (y+2)^2 = 2 --- &
Circle2: x^2 = (y-2)^2 = k --- @

Add 2 to b.s. of @ and make it equal to 2.

now, solve simultaneously...

after solving simultaneously i get:

k=6x-8y-7

Is that correct thus far?
Where do we go next?

that wouldn't be possible because there is only 2 solutions.

you have given us infinite solutions for K.

just draw the two things...

then you would almost definatly see that when k = 1 it touches it once... and if u think a little you will notice that there is a second bigger circle which will also touch the other circle once. You either see the second one through experience or intelligence.

 

[Stefano]

that wouldn't be possible because there is only 2 solutions.

you have given us infinite solutions for K.

just draw the two things...

then you would almost definatly see that when k = 1 it touches it once... and if u think a little you will notice that there is a second bigger circle which will also touch the other circle once. You either see the second one through experience or intelligence.

OH MY GOODNESS....

I just drew the 2 fken circles and the answer was jumping at me; almost poked me in the frigin eye ffs!

*sigh*... feel free to laugh.


Thanks haboozin!