If a circle passing through (1, -2) has x - y | vedclass

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(D) The center of the circle is the intersection point of its diameters $x - y = 2$ and $2x + 3y = 14$. Solving $x - y = 2$ for $y$,we get $y = x - 2$

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$5$

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(D) The center of the circle is the intersection point of its diameters $x - y = 2$ and $2x + 3y = 14$. Solving $x - y = 2$ for $y$,we get $y = x - 2$. Substituting this into the second equation: $2x + 3(x - 2) = 14$. $2x + 3x - 6 = 14$ $\Rightarrow 5x = 20$ $\Rightarrow x = 4$. Then $y = 4 - 2 = 2$. So,the center of the circle is $(4, 2)$. The circle passes through $(1, -2)$. The radius $r$ is the distance between the center $(4, 2)$ and the point $(1, -2)$. $r = \sqrt{(4 - 1)^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

If a circle passing through $(1, -2)$ has $x - y = 2$ and $2x + 3y = 14$ as its diameters,then the radius of the circle is

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