If a circle S passing through the points A(1, | vedclass

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(A) Let the centre of the circle be $C(h, k)$. Since $C$ lies in the third quadrant,$h < 0$ and $k < 0$.The line $AB$ passes through $(1, 2)$ and $(2,

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lies inside the circle $S$

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(A) Let the centre of the circle be $C(h, k)$. Since $C$ lies in the third quadrant,$h The line $AB$ passes through $(1, 2)$ and $(2, 1)$. The equation of line $AB$ is $y - 1 = \frac{2-1}{1-2}(x - 2) \Rightarrow y - 1 = -1(x - 2) \Rightarrow x + y - 3 = 0$.The distance of $C(h, k)$ from $AB$ is $\frac{|h + k - 3|}{\sqrt{1^2 + 1^2}} = \frac{7}{\sqrt{2}}$.Since $C$ is in the third quadrant,$h Since $C$ is equidistant from $A(1, 2)$ and $B(2, 1)$,$CA^2 = CB^2 \Rightarrow (h-1)^2 + (k-2)^2 = (h-2)^2 + (k-1)^2$.$h^2 - 2h + 1 + k^2 - 4k + 4 = h^2 - 4h + 4 + k^2 - 2k + 1 \Rightarrow 2h = 2k \Rightarrow h = k$.Substituting $h = k$ into $h + k = -4$,we get $2h = -4 \Rightarrow h = -2, k = -2$. So $C = (-2, -2)$.The radius $R$ is the distance $CA = \sqrt{(-2-1)^2 + (-2-2)^2} = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = 5$.The distance of $P(1, -2)$ from $C(-2, -2)$ is $\sqrt{(1 - (-2))^2 + (-2 - (-2))^2} = \sqrt{3^2 + 0^2} = 3$.Since $3 < 5$,the point $P(1, -2)$ lies inside the circle $S$.

If a circle $S$ passing through the points $A(1, 2)$ and $B(2, 1)$ has its centre $C$ located in the third quadrant at a distance of $\frac{7}{\sqrt{2}}$ units from the line $AB$,then the point $P(1, -2)$:

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