Find the point in the circle

Question

The circle $C$, given by the equation:

$$x^2 + y^2 + (1+k)y - (k+1) = 0 $$

pass through the same two points for every real number $k$.

• Find the coordinates of these two points.

• Find the minimum value of the radius of a circle $C$.

Answer 1

$x^2 + y^2 + (1+k)y - (k+1) = 0$ doesn't pass through two fixed points. If it did then let (X, Y) be one of the fixed points, so that for $k_1$ and $k_2$ then $X^2 + Y^2 + (1+k_1)Y - (k_1+1) = 0$ and $X^2 + Y^2 + (1+k_2)Y - (k_2+1) = 0$

Subtracting, then $(1+k_1)Y - (k_1+1) - (1+k_2)Y + (k_2+1) = 0$ for all $k_1, k_2$, i.e. $(k_1 - k_2)Y = (k_1 - k_2)$ for all $k \implies Y = 1$.

So far so good, but when you substitute back in the equation you have $X^2 + 1 +(k + 1) - (k + 1) = 0$, i.e. X =$ +/- i$.

Answer 2

Hint: Note you can rewrite the expression as $$x^2 + y^2 + (1+k)y - (k+1) = 0\\x^2+\left(y+\frac {k+1}{2}\right)^2-\left(\frac {k+1}{2}\right)^2-(k+1)=0\\x^2+\left(y+\frac {k+1}{2}\right)^2=(k+1)\left(\frac{k+1}{4}+1\right)\\x^2+\left(y+\frac {k+1}{2}\right)^2=(k+1)\left(\frac{k+5}{4}\right)$$ The equation obtained represents a set of circles with centre $\left(0;-\frac{k+1}{2}\right)$ and radius $\sqrt{(k+1)\left(\frac{k+5}{4}\right)}$.

Answer 3

$$x^2 + y^2 + (1+k)y - (k+1) = 0\iff x^2+y^2-1+k(y-1)=0 $$ actually implies the equation of any arbitrary circle passing through the intersection of $$x^2+y^2-1=0\ \ \ \ (1)\text{ and }y-1=0\ \ \ \ (2)$$

Setting $\displaystyle y=1$ in $\displaystyle(1),x^2=0\iff x=0$

So, the two points of intersection coincide $\displaystyle(0,1)$ which is the only fixed point.

As already found, the radius is $\displaystyle\frac{\sqrt{(k+1)(k+5)}}2$

Now the minimum positive value of $\displaystyle(k+1)(k+5)$ is clearly $0$ resulting radius $=0$