For what value of t do the four distinct poin | vedclass

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(C) Four points $(x_1, y_1), (x_2, y_2), (x_3, y_3),$ and $(x_4, y_4)$ are concyclic if the cross-ratio of their complex numbers is real,or equivalent

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(C) Four points $(x_1, y_1), (x_2, y_2), (x_3, y_3),$ and $(x_4, y_4)$ are concyclic if the cross-ratio of their complex numbers is real,or equivalently,if the determinant of the matrix formed by their coordinates is zero.Alternatively,for four points to be concyclic,the power of the point $(0, t)$ with respect to the circle passing through $(2, 3), (0, 2),$ and $(4, 5)$ must be zero.Let the circle equation be $x^2 + y^2 + 2gx + 2fy + c = 0$.Substituting $(2, 3): 4 + 9 + 4g + 6f + c = 0 \Rightarrow 4g + 6f + c = -13$.Substituting $(0, 2): 0 + 4 + 0g + 4f + c = 0 \Rightarrow 4f + c = -4$.Substituting $(4, 5): 16 + 25 + 8g + 10f + c = 0 \Rightarrow 8g + 10f + c = -41$.Subtracting the first from the third: $4g + 4f = -28 \Rightarrow g + f = -7$.From $4f + c = -4$,we have $c = -4 - 4f$.Substitute into the first: $4g + 6f - 4 - 4f = -13 \Rightarrow 4g + 2f = -9$.Solving $g+f = -7$ and $4g+2f = -9$: $2g + 2f = -14$ and $4g + 2f = -9 \Rightarrow 2g = 5 \Rightarrow g = 2.5, f = -9.5$.Then $c = -4 - 4(-9.5) = -4 + 38 = 34$.The circle is $x^2 + y^2 + 5x - 19y + 34 = 0$.For $(0, t)$ to lie on this circle: $0^2 + t^2 + 5(0) - 19(t) + 34 = 0 \Rightarrow t^2 - 19t + 34 = 0$.Factoring: $(t - 2)(t - 17) = 0$. Thus $t = 2$ or $t = 17$. Since the points must be distinct,$t = 17$ is the valid solution.

For what value of $t$ do the four distinct points $(2, 3), (0, 2), (4, 5),$ and $(0, t)$ lie on a circle?

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