If the points (2,0), (0,1), (4,5) and (0, c) | vedclass

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(B) Let the equation of the circle be $x^{2} + y^{2} + 2gx + 2fy + k = 0$. Since it passes through $(2,0)$,we have $4 + 4g + k = 0 \Rightarrow k = -4

Vedclass · Accepted Answer

$\frac{14}{3}$

Vedclass · Answer

(B) Let the equation of the circle be $x^{2} + y^{2} + 2gx + 2fy + k = 0$. Since it passes through $(2,0)$,we have $4 + 4g + k = 0 \Rightarrow k = -4 - 4g$. Since it passes through $(0,1)$,we have $1 + 2f + k = 0 \Rightarrow k = -1 - 2f$. Since it passes through $(4,5)$,we have $16 + 25 + 8g + 10f + k = 0 \Rightarrow 41 + 8g + 10f + k = 0$. Solving these equations,we get $g = -\frac{13}{6}$,$f = -\frac{17}{6}$,and $k = \frac{14}{3}$. The equation of the circle is $x^{2} + y^{2} - \frac{13}{3}x - \frac{17}{3}y + \frac{14}{3} = 0$,which simplifies to $3(x^{2} + y^{2}) - 13x - 17y + 14 = 0$. Since $(0, c)$ lies on this circle,we substitute $x=0$ and $y=c$: $3(0^{2} + c^{2}) - 13(0) - 17(c) + 14 = 0$. $3c^{2} - 17c + 14 = 0$. $(3c - 14)(c - 1) = 0$. Thus,$c = 1$ or $c = \frac{14}{3}$. Since $(0,1)$ is already a given point,the other point is $(0, \frac{14}{3})$,so $c = \frac{14}{3}$.

If the points $(2,0), (0,1), (4,5)$ and $(0, c)$ are concyclic,then the value of $c$ is

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