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Q.
The distance between the circumcentre and the centroid of the triangle formed by the vertices $(1,2),(3,-1)$ and $(4,0)$ is
TS EAMCET 2019
Solution:
We have,
$A(1,2), B(3,-1), C(4,0)$
Centroid of $\Delta A B C$ is
$\left(\frac{1+3+4}{3}, \frac{2-1+0}{3}\right)=\left(\frac{8}{3}, \frac{1}{3}\right)$
Let $(h, k)$ be circumcentre of $\Delta A B C$
$\therefore O A=O B=O C, O A=O B$
$(h-1)^{2}+(k-2)^{2}=(h-3)^{2}+(k+1)^{2}$
$\Rightarrow h^{2}-2 h+1+k^{2}-4 k+4=h^{2}-6 h+9+k^{2}+2 k+1$
$\Rightarrow 4 h-6 k=5$ ...(i)
Similarly, $O A=O C$
$\because (h-1)^{2}+(k-2)^{2}=(h-4)^{2}+k^{2}$
$\Rightarrow 6 h-4 k=11$ ...(ii)
Solving Eqs. (i) and (ii), we get
$h=\frac{23}{10}, k=\frac{7}{10}$
Distance between circumcentre and centroid of $\triangle A B C$ is
$\sqrt{\left(\frac{23}{10}-\frac{8}{3}\right)^{2}+\left(\frac{7}{10}-\frac{1}{3}\right)^{2}}=\frac{11 \sqrt{2}}{30}$
