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(B) Let the equation of the circle passing through the origin be $x^{2} + y^{2} - 2ax - 2by = 0$,where $A = (2a, 0)$ and $B = (0, 2b)$.Given $OA + 2OB

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the line $y = 2x$

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(B) Let the equation of the circle passing through the origin be $x^{2} + y^{2} - 2ax - 2by = 0$,where $A = (2a, 0)$ and $B = (0, 2b)$.Given $OA + 2OB = K$,we have $2a + 2(2b) = K$,which simplifies to $2a + 4b = K$.Substituting $2a = K - 4b$ into the circle equation:$x^{2} + y^{2} - (K - 4b)x - 2by = 0$$x^{2} + y^{2} - Kx + 4bx - 2by = 0$$(x^{2} + y^{2} - Kx) + 2b(2x - y) = 0$.For the circle to pass through a fixed point $P$ independent of $b$,the terms must satisfy $x^{2} + y^{2} - Kx = 0$ and $2x - y = 0$.From $2x - y = 0$,we get $y = 2x$.Thus,the fixed point $P$ lies on the line $y = 2x$.

$A$ variable circle is drawn passing through the origin $O$. It intersects the $X$ and $Y$ axes at points $A$ and $B$ respectively,such that $OA + 2OB = K$ (a non-zero constant). The circle always passes through a fixed point $P$ other than the origin. The point $P$ lies on -

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