Question

The radius of the circumcircle of the triangle $PRS$ is?

• A. $5$
• B. $3\sqrt{3}$
• C. $3\sqrt{2}$
• D. $2\sqrt{3}$

Answer 1

Answer: (B)

$3\sqrt{3}$

$P\equiv (1,2\sqrt{2})$

$R\equiv (9,0)$

$S\equiv (-1,0)$

Midpoint of $PR=(\frac{1+9}{2},\frac{2\sqrt{2}+0}{2})=(5,\sqrt{2})$

Equation of line $PR$ is $x+2\sqrt{2}y=9$

Therefore perpendicular bisector of $PR$ is $(y-\sqrt{2})=2\sqrt{2}(x-5)$

Midpoint of $PS=(\frac{1+(-1)}{2},\frac{2\sqrt{2}+0}{2})=(0,\sqrt{2})$

Equation of line $PS$ is $2\sqrt{2}y=4(x+1)$

Therefore, perpendicular bisector of $PS$ is $(y-\sqrt{2})=\frac{-1}{\sqrt{2}}(x-0)$

The perpendicular bisector of $PR$ and $PS$ intersect at the point $(4,-2\sqrt{2})$

This is the circumcentre of $\Delta PRS$

So, radius of circumcircle $=\sqrt{(9-4{)}^2+(0-(-2\sqrt{2}){)}^2}=3\sqrt{3}$

So, the answer is option (B).