Find the circumcentre of the triangle whose vertices are $(-2, -3), (-1, 0), (7, -6)$.

Given:

Given vertices of the triangle are $(-2, -3), (-1, 0), (7, -6)$.

To do:

We have to find the circumcentre of the given triangle.

Solution:

Let $ABC$ be a triangle whose vertices are $A (-2, -3), B (-1, 0)$ and $C (7, -6)$.

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,

\( \mathrm{OA}^{2}=(x+2)^{2}+(y+3)^{2} \)

\( =x^{2}+4 x+4+y^{2}+6 y+9 \)

\( =x^{2}+y^{2}+4 x+6 y+13 \)

\( \mathrm{OB}^{2}=(x+1)^{2}+(y+0)^{2} \)

\( =x^{2}+2 x+1+y^{2} \)

\( =x^{2}+y^{2}+2 x+1 \)

\( \mathrm{OC}^{2}=(x-7)^{2}+(y+6)^{2} \)

\( =x^{2}-14 x+49+y^{2}+12 y+36 \)

\( =x^{2}+y^{2}-14 x+12 y+85 \)

\( \mathrm{OA}^{2}=\mathrm{OB}^{2} \)

\( \Rightarrow x^{2}+y^{2}+4 x+6 y+13=x^{2}+y^{2}+2 x+1 \)

\( \Rightarrow  4 x+6 y-2 x=1-13 \)

\( \Rightarrow 2 x+6 y=-12 \)

\( \Rightarrow x+3 y=-6 \)

\( \Rightarrow x=-3y-6 \).......(i)
\( \mathrm{OB}^{2}=\mathrm{OC}^{2} \)

\( \Rightarrow x^{2}+y^{2}+2 x+1=x^{2}+y^{2}-14 x+12 y+85 \)

\( \Rightarrow 2 x+14 x-12 y=85-1 \)
\( \Rightarrow 16 x-12 y=84 \)
\( \Rightarrow 4 x-3 y=21 \)......(ii)
Substituting the value of \( x \) in (ii), we get,

\( 4(-3 y-6)-3 y=21 \)

\( \Rightarrow -12 y-24-3 y=21 \)

\( \Rightarrow -15 y=21+24 \)

\( \Rightarrow -15 y=45 \)

\( \Rightarrow y=\frac{45}{-15}=-3 \)

This implies,

 \( x=-3 y-6=-3 \times(-3)-6 \)

\( =9-6=3 \)
Therefore, the circumcentre of the triangle is $(3,-3)$.

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Updated on: 10-Oct-2022

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