Locus (1 Viewer)

Lukybear · Dec 5, 2009

A (4,0) is a point where the circle with the equation x^2+y^2=16 cuts the X-axis. Find the locus of the midpoints of all chords of this circle that contains the point A.

untouchablecuz · Dec 6, 2009

untouchablecuz · Dec 6, 2009

untouchablecuz said:
$\text{See diagram:}\\M(\frac{x+4}{2},\frac{y}{2})\\\text{We know by circle geometry that }O\hat{M}A\text{ is }\frac{\pi}{2}\\\therefore m_{OM}\times m_{MA}=-1\\\text{The locus is thus }\frac{\frac{y}{2}-0}{\frac{x+4}{4}-0} \times \frac{y-0}{x-4}=-1\\\text{i.e. } \frac{x^2}{16}+\frac{y^2}{8}=1$

gah, this is wrong. disregard everything

new solution + diagram:

$\\\text{Let the midpoint, M, of the general chord be }(x,y)\\\text{We know by circle geometry that }O\hat{M}A\text{ is }\frac{\pi}{2}\\\therefore m_{OM}\times m_{MA}=-1\\\text{The locus is thus }\frac{y-0}{x-0} \times \frac{y-0}{x-4}=-1\\\text{i.e. } (x-2)^2+y^2=2^2$

The Nomad · Dec 6, 2009

untouchablecuz · Dec 6, 2009

The Nomad said:

never seen this method before

cool

The Nomad · Dec 7, 2009

untouchablecuz said:
never seen this method before

cool

Perhaps you could tell me how to post up the answer properly like you always do...rather than uploading to imageshack and ending up with a black picture...

Locus (1 Viewer)

Lukybear

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untouchablecuz

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untouchablecuz

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The Nomad

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untouchablecuz

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The Nomad

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