The points $(2, 3), (0, 2), (4, 5)$ and $(0, t)$ are con cyclic if the value of $t$ is

2

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1

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17

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19

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Solution

The correct options are
B $2$
C $17$
Let the equation of the circle passing through $(2, 3), (0, 2)$ and $(4, 5)$ be
$x^{2} + y^{2} + 2 g x + 2 f y + c = 0 (1)$
Substituting the coordinates in (1), we get
$4 + 9 + 4 g + 6 f + c = 0 \Rightarrow 48 + 6 f + c = 13, (2)$ $0 + 4 + 0 g + 4 f + c = 0 \Rightarrow 4 f + c = - 4 (3)$
and $16 + 25 + 8 g + 10 f + c = 0$ $\Rightarrow 8 g + 10 f + c = - 41 (4)$
Solving $(2), (3)$ and $(4)$ , we get $g = \frac{5}{2}, f = - \frac{19}{2}$ and $c = 34$ , so that equation $(i)$ becomes
$x^{2} + y^{2} + 5 x - 19 y + 34 = 0 (5)$
Therefore, the point $(0, t)$ will lie on the circle $(5)$ if $t^{2} - 19 t + 34 = 0,$ i.e.. $t = 2$ or $17$

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