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

Fit the general circle $x^2+y^2+2gx+2fy+d=0$ through the three known points, then impose it on $(0,c)$ and solve.

Let the circle through $(2,0),(0,1),(4,5)$ be $x^2+y^2+2gx+2fy+d=0$.

$(2,0)$: $4+4g+d=0 \implies 4g+d=-4$ …(i)

$(0,1)$: $1+2f+d=0 \implies 2f+d=-1$ …(ii)

$(4,5)$: $16+25+8g+10f+d=0 \implies 8g+10f+d=-41$ …(iii)

From (i): $d=-4-4g$. Substituting into (ii): $2f-4-4g=-1 \implies f=\dfrac{3+4g}{2}$.

Substituting $d$ and $f$ into (iii):

$8g+10\cdot\frac{3+4g}{2}+(-4-4g)=-41 \implies 8g+15+20g-4-4g=-41 \implies 24g+11=-41 \implies g=-\frac{13}{6}.$

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

So the circle is $x^2+y^2-\dfrac{13}{3}x-\dfrac{17}{3}y+\dfrac{14}{3}=0$ (verified against all three given points).

For $(0,c)$ to lie on it:

$c^2+2fc+d=0 \implies c^2-\frac{17}{3}c+\frac{14}{3}=0 \implies 3c^2-17c+14=0.$

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