Circle equation transformations

Question

If we have a circle of the form $x^2 + y^2 = 4$ it has the middle point $(0,0)$ and radius $2$. If we want to move it $5$ to the right, the equation becomes $(x-5)^2 + y^2 = 4.$
What happens if we put a number in front of the $x$ or $y,$ eg: $5x-5?$
A graphical explanation is appreciated.
Thanks in advance.

Answer 1

Modifying the coefficient of the $x^2$ and $y^2$ will have the effect of stretching/squeezing the circle. This will result in an ellipse. For your given example, putting 5 in front of $x^2$ will give an ellipse with eccentricity $e = {\sqrt {24\over 25}}$

Answer 2

Case 1) Just replacing x by kx alone; where k is a positive constant

We end up with and equation in the form $k^2x^2 + y^2 + …. = 0$. It is an equation for an ellipse. The shape has been changed and it is NOT the “translation type of transformation”.

Case 2) Just replacing x by kx; where k is a negative constant

After applying the transformation, we get $k^2x^2 + y^2 + … = 0$. This means the result curve is again an ellipse and similar comment as the above applies.

Case 3) Replacing x by kx and y by ky simultaneously; where k is a non-zero constant

This is better explained using its general form ($x^2 + y^2 + Dx + Ey + F = 0$) instead. After applying the transformation, we get $(kx)^2 + (ky)^2 + Dkx + Eky + F = 0$

i.e. $x^2 + y^2 + (\dfrac {D}{k})x + (\dfrac {E}{k})y + (\dfrac {F}{k^2}) = 0$

The resultant equation represents a circle but with the center shifted and the radius changed.