If $A(-2,1), B(0,-2), C(1,2)$ are the vertices of a triangle $A B C$, then the perpendicular distance from its circumcentre to the side $B C$ is

• A

  $\frac{7 \sqrt{13}}{22}$
• Correct answer
  B

  $\frac{3 \sqrt{17}}{22}$
• C

  $\frac{5 \sqrt{10}}{11}$
• D

  $\frac{\sqrt{2026}}{22}$

Solution

Let $(h, k)$ be circumcentre of $\triangle A B C$ and $A(-2,1)$, $B(0,-4, C(1,4)$ are the vertices of $\triangle A B C$.
$$ \begin{array}{ll} \therefore \quad & O A=O B=O C \\ & O A=O B \\ & (h+2)^2+(k-1)^2=h^2+(k+2)^2 \\ \Rightarrow \quad & 4 h-6 k+1=0 \end{array} $$ $$ \begin{aligned} & O B=O C \\ & h^2+(k+2)^2=(h-1)^2+(k-2)^2 \\ & =2 h+8 k-1=0 \end{aligned} $$ On solving Eqs. (i) and (ii), we get $$ h=-\frac{1}{22}, k=\frac{3}{22} $$ Equation of $B C$ $$ \begin{aligned} & y+2=\frac{2+2}{1-0}(x) \\ & 4 x-y-2=0 \end{aligned} $$ Distance from circumcentre to line $B C$ $$ \begin{aligned} & d=\left|\frac{4\left(\frac{-1}{22}\right)-\frac{3}{22}-2}{\sqrt{4^2+1^2}}\right| \\ & d=\frac{51}{22 \sqrt{17}}=\frac{3 \sqrt{17}}{22} \end{aligned} $$

Hence, the correct answer is option: 'B'.