# 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.

Answer

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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.

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.

Last updated date: 01st Oct 2023

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