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# Find the coordinates of the circumcentre of the triangle whose vertices are $\left( {8,6} \right),\left( {8, - 2} \right)$ and$PA = PB = PC = \sqrt {{{\left( {5 - 8} \right)}^2} + {{\left( {2 - 6} \right)}^2}} = 5$$\left( {2, - 2} \right)$. Also, find its circum-radius.

Last updated date: 19th Sep 2024
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Hint: A circumcentre of a triangle is equidistant from all the vertices of the triangle.

Keeping in mind the above point, let us consider $A\left( {8,6} \right),B\left( {8, - 2} \right)$ and $C\left( {2, - 2} \right)$ as the vertices of the given triangle. If we consider a point $P\left( {x,y} \right)$ which is equidistant from all three vertices,
Then,
$PA = PB = PC$
Now we’ll use the formula to find the length of $PA$, $PB$ and $PC$ equate them, to simplify the calculations we are going to square the equations.
Therefore,
$\Rightarrow P{A^2} = P{B^2} = P{C^2}$
Now let us take the first two terms and calculate the length of the two lines,
Therefore,
$\Rightarrow P{A^2} = P{B^2}$
Now we’ll use the formula to find the length of lines if $R\left( {{a_2},{b_2}} \right),S\left( {{a_1},{b_1}} \right)$ are the points:
$RS = \sqrt {{{\left( {{a_1} - {a_2}} \right)}^2} + {{\left( {{b_1} - {b_2}} \right)}^2}}$
Therefore if we apply the above formula, we get,
${\left( {x - 8} \right)^2} + {\left( {y - 6} \right)^2} = {\left( {x - 8} \right)^2} + {\left( {y + 2} \right)^2}$

On further solving the above equation, we get,
${x^2} + {y^2} - 16x - 12y + 100 = {x^2} + {y^2} - 16x + 4y + 68$
$\Rightarrow 16y = 32$
$\Rightarrow y = 2$
Same steps are to be performed for the other two points and $PC$,
$P{B^2} = P{C^2}$
${\left( {x - 8} \right)^2} + {\left( {y + 2} \right)^2} = {\left( {x - 2} \right)^2} + {\left( {y + 2} \right)^2}$
On solving further, we get,
${x^2} + {y^2} - 16x + 4y + 68 = {x^2} + {y^2} - 4x + 4y + 8$
$\Rightarrow 12x = 60$
$\Rightarrow x = 5$
We have found the values of $x$ and$y$.
So the circumcentre, will be equal to $\left( {5,2} \right)$
To find the circum-radius,
$PA = PB = PC = \sqrt {{{\left( {5 - 8} \right)}^2} + {{\left( {2 - 6} \right)}^2}} = 5$

Note: Remember the fact that a circumcentre of a triangle is equidistant from all the vertices of the triangle which can be used to arrive at the solution.