Question

prove geometrically, if the distances of a point from two fixed points are in constant ratio, and the ratio is not equal to 1, then the locus of the variable point will be a circle

Answer 1

This is in two dimeensional space right?

Answer 2

yeah

Answer 3

I got it.

Answer 4

Isn't it supposed to be an ellipse?

Answer 5

i dont think that

Answer 6

in ellipse, the sum of distances of any point on it from the foci are constant

Answer 7

in fact, if u take 2 fixed points and write the equation for any arbitrary point satisfying the property, u will get an equation of a circle

Answer 8

Suppose the fixed points are A, at (x_1,y_1), and B, at (x_2, y_2). We're given that the distance of point P, at (x,y), from points A and B is in a fixed ratio, i.e. AP/BP = k, where k is not 1. We can say then that
\[AP=kBP\]
AP's length is
\[\sqrt{(x-x_1)^2+(y-y_1)^2}\]
and BP's length is
\[\sqrt{(x-x_2)^2+(y-y_2)^2}\]
So we have
\begin{eqnarray*}
\sqrt{(x-x_1)^2+(y-y_1)^2}&=&k\sqrt{(x-x_2)^2+(y-y_2)^2} \\
(x-x_1)^2+(y-y_1)^2&=&k^2((x-x_2)^2+(y-y_2)^2) \\
x^2-2xx_1+x_1^2+y_2-2yy_1+y_1^2&=&k^2(x^2-2xx_2+x_2^2+y^2-2yy_2+y_2^2) \\
x^2-2xx_1+x_1^2+y_2-2yy_1+y_1^2&=&k^2x^2-2k^2xx_2+k^2x_2^2+k^2y^2-2k^2yy_2+k^2y_2^2
\end{eqnarray*}
to be continued

Answer 9

i solved by that method, but i wanna know the synthetic geometrical proof.

Answer 10

so you know the answer by this method and you want a construction diagram?

Answer 11

yes, exactly. can u help me, please?

Answer 12

constructing this would be a bit difficult seeing as the radius of the resulting circle is quite a complicated expression, and so are the coordinates of the centre of the circle.

Answer 13

ok, i will b happy if anyone answers this. thanx 4 ur co-operation

Answer 14

if you want I can finish my algebraic argument. It has a subtle trick at a certain step.

Answer 15

certainly, you can do. i can verify my process

Answer 16

Part 2:
\[(1-k^2)x^2-2x(x_1-k^2x_2)+(x_1^2-k^2x_2^2)+(1-k^2)y_2^2-2y(y_1-k^2y_2)+(y_1^2-k^2y_2^2)=0\]
\[x^2-\frac{2(x_1-k^2x_2)}{1-k^2}+\frac{x_1^2-k^2x_2^2}{1-k^2}+y^2-\frac{2(y_1-k^2y_2)}{1-k^2}+\frac{y_1^2-k^2y_2^2}{1-k^2}=0\]
We consider
\[x^2-\frac{2(x_1-k^2x_2)}{1-k^2}+\frac{x_1^2-k^2x_2^2}{1-k^2}\]
only; the other three terms would be transformed in exactly the same way. Completing the square of the term, we get:
\[x^2-\frac{2(x_1-k^2x_2)}{1-k^2}+\Big(\frac{x_1-k^2x_2}{1-k^2}\Big)^2+\frac{x_1^2-k^2x_2^2}{1-k^2}-\Big(\frac{x_1-k^2x_2}{1-k^2}\Big)^2\]
\[=\Big(x-\frac{x_1-k^2x_2}{1-k^2}\Big)^2+\Big(\frac{x_1^2-k^2x_2^2}{1-k^2}-\Big(\frac{x_1-k^2x_2}{1-k^2}\Big)^2\Big)\]
Combining this with the other one, we get
\[(x-a)^2+(y-b)^2=r^2\]
where
\begin{eqnarray*}
a&=&\frac{x_1-k^2x_2}{1-k^2} \\
b&=&\frac{y_1-k^2y_2}{1-k^2} \\
r^2&=&\Big(\frac{x_1^2-k^2x_2^2}{k^2-1}-\Big(\frac{x_1-k^2x_2}{k^2-1}\Big)^2\Big)+\Big(\frac{y_1^2-k^2y_2^2}{k^2-1}-\Big(\frac{y_1-k^2y_2}{k^2-1}\Big)^2\Big)
\end{eqnarray*}
Notice also that if k=1, the denominator would vanish, which would make the whole argument above fail.