Q.If (2, 0), (0, 1), (4, 5) and (0, c) are concyclic then find c.

Fit a general circle through the three known points to fix $g,f,k$; the fourth point $(0,c)$ then gives a quadratic in $c$ whose roots are the two $y$-intercepts of the circle.

Let the circle be $x^2+y^2+2gx+2fy+k=0$.

Substituting $(2,0)$: $4+4g+k=0$ ... (i)

Substituting $(0,1)$: $1+2f+k=0$ ... (ii)

Substituting $(4,5)$: $41+8g+10f+k=0$ ... (iii)

From (i): $k=-4-4g$. Substitute in (ii): $2f=3+4g \Rightarrow f=\dfrac{3+4g}{2}$

Substitute into (iii): $24g+11=-41 \Rightarrow g=-\dfrac{13}{6}$

Then $f=\dfrac{3+4(-13/6)}{2}=-\dfrac{17}{6}$, and $k=-4-4\left(-\dfrac{13}{6}\right)=\dfrac{14}{3}$

Now substitute $(0,c)$: $c^2+2fc+k=0$ …