Locus Question and derivatives graphs (1 Viewer)

[Loz_metalhead]

My teacher is going to be away all week and my math exam is next tuesday.
We havent even done the locus chapter in the book and he spent one lesson on it...its going to be in the exam. I sort of get it but dont know how to do this question

A point P(x,y) moves so that its distance from the point Q(-2,8) is twice its distance from the point R(10,-1). Show that the locus of P is a circle and write down its cetre and radius.

I drew a diagram and tried to work it out but I have no idea. The answer is:
(14,-4), 10 units.

Also, my teacher spent one lesson explaining derivatives graphs and this was the lesson before an exam last term...no-one in our class got it right. I really dont understand how to do it and he wont be here all week ...

 

[Trev]

Since PQ is twice the distance from PR then PQ=2PR.
√[(x+2)²+(x-8)²]=2√[(x-10)²+(y+1)²]
[(x+2)²+(x-8)²]=4[(x-10)²+(y+1)²]... etc.
Expand and simplify.

 

[Loz_metalhead]

I have expanded and simplified but I dont know how to work out the y and x values...and r.

3(x^2 + 112- 28x + y^2 + 8y)=0

 

[Trev]

x²-28x+y²+8y+112=0
(x-14)²-14²+(y+4)²-4²+112=0
(x-14)²+(y+4)²=100
So centre circle is (12,-4) with radius 10.

 

[Loz_metalhead]

Thankyou very much. :wave:

 

[SoulSearcher]

x²-28x+y²+8y+112=0
(x-14)²-14²+(y+4)²-4²+112=0
(x-14)²+(y+4)²=100
So centre circle is (12,-4) with radius 10.

(14, -4) isn't it?

 

[Trev]

Yeah sorry, typo.