Finding the point through which a circle passes if the general equation is given

Question

No idea how it is even possible to do this.

So we have the general equation (but the y-coordinate of the center is still kept variable as you can see below):- $$x^2+y^2-2x-3ky-2=0$$

We have to find through which two fixed points does the circle pass.

Answer 1

Completing the squares you see that the circles are of the form $$ (x-1)^2 + (y-\frac{3}{2}k)^2 = 3 + \frac{9}{4}k^2$$ So we have a family of circles with center $(1,\frac{3}{2}k)$ and radius $\sqrt{3 + \frac{9}{4}k^2}$ By symmetry, if the circles have exactly two points in common, this has to happen for $y=0$.

In fact, you will find exactly two values of $x$ wich solve the equation you get if $y=0$. I won't show you, though, why these are the only solutions...

Answer 2

Hint: at which points does your circle meet the axes?

Answer 3

Hint If the points $(x, y)$ of interest satisfy the equation for distinct parameter values $k$ and $k'$, then subtracting the two equations gives that $(x, y)$ must satisfy $$3ky' - 3ky = 0.$$