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Question

# The circumcentre of the triangle with vertices $\left(8,6\right),\left(8,-2\right)$ and $\left(2,-2\right)$ is at the point

A

$\left(2,-1\right)$

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B

$\left(1,-2\right)$

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C

$\left(5,2\right)$

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D

$\left(2,5\right)$

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E

$\left(4,5\right)$

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Solution

## The correct option is C $\left(5,2\right)$Explanation of the correct option:Finding the circumcentre of the triangle:Step 1: Equating the circumradius of triangle to find the value of $y$Let $A\left(8,6\right),B\left(8,-2\right)$ and $C\left(2,-2\right)$ be the vertices of a triangle. Let $P\left(x,y\right)$ be circumcentre of the triangle. Then, $\mathrm{PA}=\mathrm{PB}=\mathrm{PC}$Now, ${\mathrm{PA}}^{2}={\mathrm{PB}}^{2}$.$\therefore {\left(x-8\right)}^{2}+{\left(y-6\right)}^{2}={\left(x-8\right)}^{2}+{\left(y+2\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(y-6\right)}^{2}={\left(y+2\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{y}^{2}-12y+36={y}^{2}+4y+4\phantom{\rule{0ex}{0ex}}⇒16y=32\phantom{\rule{0ex}{0ex}}⇒y=2$Step 2: Equating the circumradius of the triangle to find the value of $x$Similarly, ${\mathrm{PB}}^{2}={\mathrm{PC}}^{2}$.$\therefore {\left(x-8\right)}^{2}+{\left(y+2\right)}^{2}={\left(x-2\right)}^{2}+{\left(y+2\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-8\right)}^{2}={\left(x-2\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-16x+64={x}^{2}-4x+4\phantom{\rule{0ex}{0ex}}⇒12x=60\phantom{\rule{0ex}{0ex}}⇒x=5$Hence, the coordinates of the circumcenter of a triangle is $\left(5,2\right)$.Therefore, option (C) is the correct answer.

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