The points (1,0), (0,1), (0,0) and (2k, 3k), | vedclass

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(D) The equation of the circle passing through the points $(1,0), (0,1)$ and $(0,0)$ is given by the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.Subs

Vedclass · Accepted Answer

$\frac{5}{13}$

Vedclass · Answer

(D) The equation of the circle passing through the points $(1,0), (0,1)$ and $(0,0)$ is given by the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.Substituting $(0,0)$,we get $c = 0$.Substituting $(1,0)$,we get $1 + 2g = 0 \Rightarrow g = -\frac{1}{2}$.Substituting $(0,1)$,we get $1 + 2f = 0 \Rightarrow f = -\frac{1}{2}$.Thus,the equation of the circle is $x^2 + y^2 - x - y = 0$.Since the point $(2k, 3k)$ lies on this circle,it must satisfy the equation:$(2k)^2 + (3k)^2 - (2k) - (3k) = 0$$4k^2 + 9k^2 - 5k = 0$$13k^2 - 5k = 0$$k(13k - 5) = 0$Given $k 
eq 0$,we have $13k - 5 = 0$,which implies $k = \frac{5}{13}$.

The points $(1,0), (0,1), (0,0)$ and $(2k, 3k), k \neq 0$ are concyclic,if $k$ is

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