For what value of k,the points (0, 0), (1, 3) | vedclass

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(B) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since the circle passes through $(0, 0)$,we have $c = 0$.Passing through $(1, 3)

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$1$

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(B) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since the circle passes through $(0, 0)$,we have $c = 0$.Passing through $(1, 3)$: $1^2 + 3^2 + 2g(1) + 2f(3) = 0$ $\Rightarrow 10 + 2g + 6f = 0$ $\Rightarrow g + 3f = -5$ (Equation $1$).Passing through $(2, 4)$: $2^2 + 4^2 + 2g(2) + 2f(4) = 0$ $\Rightarrow 20 + 4g + 8f = 0$ $\Rightarrow g + 2f = -5$ (Equation $2$).Subtracting Equation $2$ from Equation $1$: $(g + 3f) - (g + 2f) = -5 - (-5) \Rightarrow f = 0$.Substituting $f = 0$ into Equation $2$: $g + 2(0) = -5 \Rightarrow g = -5$.The equation of the circle is $x^2 + y^2 - 10x = 0$.For the point $(k, 3)$ to be on the circle,it must satisfy the equation:$k^2 + 3^2 - 10(k) = 0 \Rightarrow k^2 - 10k + 9 = 0$.Factoring the quadratic: $(k - 1)(k - 9) = 0$.Thus,$k = 1$ or $k = 9$. Given the options,$k = 1$ is the correct value.

For what value of $k$,the points $(0, 0), (1, 3), (2, 4)$ and $(k, 3)$ are concyclic?

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