Locus Question - Cambridge (Need Assistance) (1 Viewer)

[Pinchy444]

Hey,

If anybody could please assist me with this problem {}, it would be greatly appreciated.

The answer is: x^2+y^2=4

I'm close but can't quite get that answer.

Thanks!

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=d_{PA} = \sqrt{(4-x)^2@plus;y^2} \\~~~~~d_{PB} = \sqrt{(1-x)^2 @plus; y^2} \\\\ \sqrt{(4-x)^2@plus;y^2} =2\sqrt{(1-x)^2 @plus; y^2} \\\\ \sqrt{16 -8x@plus;x^2@plus;y^2} = 2\sqrt{1-2x@plus;x^2@plus;y^2} \\\\ 16-8x@plus;x^2@plus;y^2 = 4(1-2x@plus;x^2@plus;y^2)\\\\ 16-8x@plus;x^2@plus;y^2=4-8x@plus;4x^2@plus;4y^2 \\\\ 3x^2 @plus; 3y^2 = 12 \\\\ x^2 @plus; y^2 = 4 \\\\ \therefore ~Locus~ of ~P ~is~ x^2 @plus; y^2 = 4" target="_blank"><img src="http://latex.codecogs.com/gif.latex?d_{PA} = \sqrt{(4-x)^2+y^2} \\~~~~~d_{PB} = \sqrt{(1-x)^2 + y^2} \\\\ \sqrt{(4-x)^2+y^2} =2\sqrt{(1-x)^2 + y^2} \\\\ \sqrt{16 -8x+x^2+y^2} = 2\sqrt{1-2x+x^2+y^2} \\\\ 16-8x+x^2+y^2 = 4(1-2x+x^2+y^2)\\\\ 16-8x+x^2+y^2=4-8x+4x^2+4y^2 \\\\ 3x^2 + 3y^2 = 12 \\\\ x^2 + y^2 = 4 \\\\ \therefore ~Locus~ of ~P ~is~ x^2 + y^2 = 4" title="d_{PA} = \sqrt{(4-x)^2+y^2} \\~~~~~d_{PB} = \sqrt{(1-x)^2 + y^2} \\\\ \sqrt{(4-x)^2+y^2} =2\sqrt{(1-x)^2 + y^2} \\\\ \sqrt{16 -8x+x^2+y^2} = 2\sqrt{1-2x+x^2+y^2} \\\\ 16-8x+x^2+y^2 = 4(1-2x+x^2+y^2)\\\\ 16-8x+x^2+y^2=4-8x+4x^2+4y^2 \\\\ 3x^2 + 3y^2 = 12 \\\\ x^2 + y^2 = 4 \\\\ \therefore ~Locus~ of ~P ~is~ x^2 + y^2 = 4" /></a>

 

[Pinchy444]

Thanks for that! Much appreciated!