A circle passes through the points (2,0) and | vedclass

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(B) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the points $(2,0)$ and $(1,2)$ lie on the circle: $4 + 4g + c = 0 \Righta

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$\sqrt{\frac{5}{2}}$

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(B) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the points $(2,0)$ and $(1,2)$ lie on the circle: $4 + 4g + c = 0 \Rightarrow 4g + c = -4$  $(i)$ $1 + 4 + 2g + 4f + c = 0 \Rightarrow 2g + 4f + c = -5$  $(ii)$ The power of the point $(0,2)$ with respect to the circle is given by $S(0,2) = 0^2 + 2^2 + 2g(0) + 2f(2) + c = 4 + 4f + c$. Given the power is $4$,we have $4 + 4f + c = 4 \Rightarrow 4f + c = 0$  $(iii)$ Subtracting $(iii)$ from $(ii)$: $(2g + 4f + c) - (4f + c) = -5 - 0$ $\Rightarrow 2g = -5$ $\Rightarrow g = -\frac{5}{2}$. Substituting $g$ into $(i)$: $4(-\frac{5}{2}) + c = -4$ $\Rightarrow -10 + c = -4$ $\Rightarrow c = 6$. Substituting $c$ into $(iii)$: $4f + 6 = 0 \Rightarrow f = -\frac{6}{4} = -\frac{3}{2}$. The radius $r = \sqrt{g^2 + f^2 - c} = \sqrt{(-\frac{5}{2})^2 + (-\frac{3}{2})^2 - 6} = \sqrt{\frac{25}{4} + \frac{9}{4} - 6} = \sqrt{\frac{34}{4} - 6} = \sqrt{\frac{17}{2} - 6} = \sqrt{\frac{5}{2}}$.

$A$ circle passes through the points $(2,0)$ and $(1,2)$. If the power of the point $(0,2)$ with respect to this circle is $4$,then the radius of the circle is

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