The point P is equidistant from A(1, 3),B(-3, | vedclass

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(D) Let the point $P$ be $(x, y)$. Since $P$ is equidistant from $A, B,$ and $C$,$P$ is the circumcenter of $	riangle ABC$.$PA^2 = PB^2 = PC^2$.$PA^2

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$5\sqrt{10}$

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(D) Let the point $P$ be $(x, y)$. Since $P$ is equidistant from $A, B,$ and $C$,$P$ is the circumcenter of $	riangle ABC$.$PA^2 = PB^2 = PC^2$.$PA^2 = (x-1)^2 + (y-3)^2 = x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 + y^2 - 2x - 6y + 10$.$PB^2 = (x+3)^2 + (y-5)^2 = x^2 + 6x + 9 + y^2 - 10y + 25 = x^2 + y^2 + 6x - 10y + 34$.$PC^2 = (x-5)^2 + (y+1)^2 = x^2 - 10x + 25 + y^2 + 2y + 1 = x^2 + y^2 - 10x + 2y + 26$.Equating $PA^2 = PB^2$:$-2x - 6y + 10 = 6x - 10y + 34$ $\Rightarrow 8x - 4y + 24 = 0$ $\Rightarrow 2x - y + 6 = 0$ .....$(i)$.Equating $PA^2 = PC^2$:$-2x - 6y + 10 = -10x + 2y + 26$ $\Rightarrow 8x - 8y - 16 = 0$ $\Rightarrow x - y - 2 = 0$ .....$(ii)$.Subtracting $(ii)$ from $(i)$:$(2x - y + 6) - (x - y - 2) = 0$ $\Rightarrow x + 8 = 0$ $\Rightarrow x = -8$.Substituting $x = -8$ in $(ii)$:$-8 - y - 2 = 0 \Rightarrow y = -10$.Thus,$P = (-8, -10)$.$PA = \sqrt{(-8 - 1)^2 + (-10 - 3)^2} = \sqrt{(-9)^2 + (-13)^2} = \sqrt{81 + 169} = \sqrt{250} = 5\sqrt{10}$.

List-$I$ (Defects)	List-$II$ (Shown by)
$a$. Frenkel defect	$i$. Non-ionic solids and density of the solid decreases
$b$. Schottky defect	$ii$. Non-ionic solids and density of the solid increases
$c$. Vacancy defect	$iii$. Ionic solids and density of the solid decreases
$d$. Interstitial defect	$iv$. Ionic solids and density of the solid remains constant

The point $P$ is equidistant from $A(1, 3)$,$B(-3, 5)$,and $C(5, -1)$. Then $PA =$

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