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Find the coordinates of the circumcentre of the triangle whose vertices are $(3, 0), (-1, -6)$ and $(4, -1)$. Also, find its circumradius.
Given:
Given points are $(3, 0), (-1, -6)$ and $(4, -1)$.
To do:
We have to find the coordinates of the circumcentre of the triangle and its circumradius.
Solution:
Let $ABC$ be a triangle whose vertices are $A (3, 0), B (-1, -6)$ and $C (4, -1)$.
Let \( O(x, y) \) be the circumcentre of the \( \Delta A B C \).
This implies,
\( \mathrm{OA}=\mathrm{OB}=\mathrm{OC} \)
\( \Rightarrow \mathrm{OA}^{2}=\mathrm{OB}^{2}=\mathrm{OC}^{2} \)
We know that,
The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).
Therefore,
\( \mathrm{OA}^{2}= (x-3)^{2}+(y-0)^{2} \)
\( =(x-3)^{2}+y^{2} \)
\( \mathrm{OB}^{2}=(x+1)^{2}+(y+6)^{2} \)
\( \mathrm{O C}^{2}=(x-4)^{2}+(y+1)^{2} \)
\( \mathrm{OA}^{2}=\mathrm{OB}^{2} \)
\( (x-3)^{2}+y^{2}=(x+1)^{2}+(y+6)^{2} \)
\( \Rightarrow x^{2}-6 x+9+y^{2}=x^{2}+2 x+1+y^{2}+12 y+36 \)
\( \Rightarrow-6 x-2 x-12 y=1+36-9 \)
\( \Rightarrow-8 x-12 y=28 \)
\( \Rightarrow 2 x+3 y=-7 \).......(i)
\( \mathrm{OB}^{2}=\mathrm{OC}^{2} \)
\( (x+1)^{2}+(y+6)^{2}=(x-4)^{2}+(y+1)^{2} \)
\( x^{2}+2 x+1+y^{2}+12 y+36=x^{2}-8 x+16+\)
\( y^{2}+2 y+1 \)
\( 2 x+12 y+37+8 x-2 y=17 \)
\( 10 x+10 y=17-37=-20 \)
\( x+y=-2 \)......(ii)
Multiply (i) by 1 and (ii) by 2 and subtracting, we get,
\( (2 x+3 y)-(2x+2y)=-7-(-4) \)
\( y=-3 \)
\( \Rightarrow x-3=-2 \)
\( \Rightarrow x=-2+3=1 \)
The coordinates of \( \mathrm{O} \) are \( (1,-3) \).
Radius \( =\mathrm{OA}=\sqrt{(1+1)^{2}+(-3+6)^{2}} \)
\( =\sqrt{(2)^{2}+(3)^{2}} \)
\( =\sqrt{4+9} \)
\( =\sqrt{13} \) units
The coordinates of the circumcentre of the triangle are $(1, -3)$ and the circumradius is $\sqrt{13}$ units.
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