If (1, 1), (-2, 2), (2, -2) are 3 points on a | vedclass

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(A) Let the points be $A(1, 1), B(-2, 2)$,and $C(2, -2)$.Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Substituting the points int

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$\frac{1}{2}$

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(A) Let the points be $A(1, 1), B(-2, 2)$,and $C(2, -2)$.Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Substituting the points into the equation:For $(1, 1): 1 + 1 + 2g + 2f + c = 0 \Rightarrow 2g + 2f + c = -2$ $(i)$For $(-2, 2): 4 + 4 - 4g + 4f + c = 0 \Rightarrow -4g + 4f + c = -8$ $(ii)$For $(2, -2): 4 + 4 + 4g - 4f + c = 0 \Rightarrow 4g - 4f + c = -8$ $(iii)$Subtracting $(iii)$ from $(ii): -8g + 8f = 0 \Rightarrow g = f$.Substituting $g = f$ in $(i)$ and $(ii): 4g + c = -2$ and $-4g + 4g + c = -8 \Rightarrow c = -8$.Substituting $c = -8$ in $4g + c = -2: 4g - 8 = -2$ $\Rightarrow 4g = 6$ $\Rightarrow g = \frac{3}{2}$.Thus,$g = f = \frac{3}{2}$ and $c = -8$.The centre of the circle is $(-g, -f) = \left(-\frac{3}{2}, -\frac{3}{2}\right)$.The perpendicular distance from $\left(-\frac{3}{2}, -\frac{3}{2}\right)$ to the line $3x - 4y + 1 = 0$ is given by $d = \frac{|3(-\frac{3}{2}) - 4(-\frac{3}{2}) + 1|}{\sqrt{3^2 + (-4)^2}}$.$d = \frac{|-\frac{9}{2} + 6 + 1|}{5} = \frac{|\frac{5}{2}|}{5} = \frac{1}{2}$.

If $(1, 1), (-2, 2), (2, -2)$ are $3$ points on a circle $S$,then the perpendicular distance from the centre of the circle $S$ to the line $3x - 4y + 1 = 0$ is

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