The radius of the circumcircle of the triangle $P R S$ is\

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Professor PC Thomas Classes · Answer

(b) : Radius of circumcircle of $	riangle P R S$ Equation of perpendicular bisector of $S R$ is $x=4$\ Equation of perpendicular bisector of $P S$ is\ $y-\sqrt{2}=\frac{-1}{\sqrt{2}}(x-0)$ or $\sqrt{2} y+x=2$\ Circumcentre is point of intersection of (1) and (2),\ $x=4, y=-\sqrt{2} \quad 	herefore C(4,-\sqrt{2})$\ $	herefore \quad$ radius $=P C=\sqrt{(3)^{2}+(3 \sqrt{2})^{2}}=3 \sqrt{3}$ units\