Is the locus of a point with the ratio of its distances from two given points being a constant $\neq 1$ always a circle? Here is Prob. 29, in the Additional Problems For Chapter 1, in the book Calculus With Analytic Geometry by George F. Simmons, 2nd edition:  A point moves in such a way that the ratio of its distances from two fixed points is a constant $k \neq 1$. Show that the locus is a circle.  My Attempt:  Let the two fixed points be $Q = (a, b)$ and $R = (c, d)$. If $P = (x, y)$ is any point such that $$ \frac{PQ}{PR} = k, $$ then upon squaring and multiplying out by the denominator we obtain $$ k^2 (x-c)^2 + k^2 (y-d)^2 = (x-a)^2 + (y-b)^2, $$ which we can also rewrite as $$ \left( k^2-1\right) x^2 + \left( k^2-1\right) y^2 - 2\left( k^2 c - a \right) x - 2 \left( k^2 d - b \right) y = a^2 + b^2 -  k^2 \left( c^2 + d^2 \right), $$ and since $k &gt; 0$ and $k \neq 1$, therefore we have $k^2 -1 \neq 0$, and thus upon dividing both sides by $k^2-1$, we obtain $$ x^2 + y^2 - \frac{ 2\left( k^2 c - a \right)  }{ k^2 - 1} x - \frac{ 2 \left( k^2 d - b \right)  }{k^2-1} y = \frac{ a^2 + b^2 -  k^2 \left( c^2 + d^2 \right) }{ k^2-1},  $$ and upon collecting together the terms in $x$ and the terms in $y$, we get $$ x^2 - 2 \frac{ \left( k^2 c - a \right)  }{ k^2 - 1} x  + y^2 - 2 \frac{  \left( k^2 d - b \right)  }{k^2-1} y = \frac{ a^2 + b^2 -  k^2 \left( c^2 + d^2 \right) }{ k^2-1},  $$ and upon completing the two squares on the left-hand-side, we get $$ \begin{align}  &amp; \ \ \ \left( x - \frac{ \left( k^2 c - a \right)  }{ k^2 - 1}  \right)^2 + \left( y - \frac{  \left( k^2 d - b \right)  }{k^2-1} \right)^2 \\  &amp;= \frac{ \left( k^2 c - a \right)^2  }{ \left( k^2 - 1 \right)^2 }  + \frac{  \left( k^2 d - b \right)^2   }{ \left( k^2-1 \right)^2 } + \frac{ a^2 + b^2 -  k^2 \left( c^2 + d^2 \right) }{ k^2-1} \\ &amp;= \frac{ \left( k^2 c - a \right)^2  +  \left( k^2 d - b \right)^2   + \left( k^2-1 \right) \left( a^2 + b^2 \right) -  k^2 \left( k^2-1 \right) \left( c^2 + d^2 \right)   }{ \left( k^2 - 1 \right)^2 } \\ &amp;= \frac{  k^2 \left( 2c^2 + 2d^2 -ac - bd + a^2 + b^2  \right) - k^4 \left( c^2 + d^2 \right) }{ \left( k^2 - 1 \right)^2 } \\ &amp;= \ldots. \end{align} $$ Now the locus of the point $P = (x, y)$ is a circle if and only if $$ \frac{  k^2 \left( 2c^2 + 2d^2 -ac - bd + a^2 + b^2  \right) - k^4 \left( c^2 + d^2 \right) }{ \left( k^2 - 1 \right)^2 } &gt; 0, $$ which is the case if and only if $$ k^2 \left( 2c^2 + 2d^2 -ac - bd + a^2 + b^2  \right) - k^4 \left( c^2 + d^2 \right) &gt; 0. $$  Is what I have done so far correct in each and every detail? If so, how can we show that the last inequality always holds? Or, can we show this for all possible $a, b, c, d, k \in \mathbb{R}$ such that (i) $(a, b) \neq (c, d)$,  and (ii) $k &gt; 0$ and $k \neq 1$? Or, are there any exceptions?
Equation of a line maintaining equal ratio distance between two points. Say I have two points. One at $A = (0, 0)$ and another at $B = (0, 10)$. I wish to derive an equation for a line that, for any point $P$ on the line, would equal a set ratio for the distance $P$ to $A$ and $P$ to $B$.  This is easy if the predetermined ratio for $A:B$ is $1:1$. I would create a line from $A$ to $B$, find its midpoint, then create an orthogonal line at the midpoint. Then at any $P$ on the line I know that the ratio distances from $P$ to $A$ and $B$ will be $1:1$.  But what if the ratio for $A:B$ is something like $11:18$?  This is for a positioning algorithm I am implementing - not for homework :)  EDIT: Note - If an exact formula is not possible, an approximation of the line out to $\pm 200$ from $A$ or $B$ is acceptable as values further from $A$ and $B$ will not be utilized anyway. A linear equation for the final result/approximation is ideal for speed in running multiple computations.
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