A circle passing through (0,0), (2,6), and (6 | vedclass

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(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the circle passes through the origin $(0,0)$, we have $c = 0$. Since it p

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$5$

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(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the circle passes through the origin $(0,0)$, we have $c = 0$. Since it passes through $(2,6)$, we have $4 + 36 + 4g + 12f = 0$, which simplifies to $g + 3f = -10$. Since it passes through $(6,2)$, we have $36 + 4 + 12g + 4f = 0$, which simplifies to $3g + f = -10$. Solving the system of equations $g + 3f = -10$ and $3g + f = -10$, we subtract the equations to get $2g - 2f = 0$, so $g = f$. Substituting $g = f$ into $g + 3f = -10$, we get $4g = -10$, so $g = -2.5$ and $f = -2.5$. The equation of the circle is $x^2 + y^2 - 5x - 5y = 0$. To find the intersection with the $x$-axis, set $y = 0$: $x^2 - 5x = 0$, which gives $x(x - 5) = 0$. Thus, the points of intersection are $(0,0)$ and $(5,0)$. Since $P 
eq (0,0)$, the point $P$ is $(5,0)$. The length of $OP$ is the distance between $(0,0)$ and $(5,0)$, which is $5$.

$A$ circle passing through $(0,0)$, $(2,6)$, and $(6,2)$ cuts the $x$-axis at the point $P \neq (0,0)$. Then, the length of $OP$, where $O$ is the origin, is

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