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Hint: A circle on which the three vertices of a triangle lie is called the circumcircle of the triangle and the centre of this circle is called the circumcentre.

The given triangle is \[\Delta ABC\].

We will consider \[\left( h,k \right)\] to be coordinates of one of the vertices .

First , we need to find the distance of \[\left( h,k \right)\] from \[(1,0)\]and \[(-1,0)\].

We know that the distance between the two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] given as \[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]

So , to find the distance of \[\left( h,k \right)\] from \[(1,0)\], we will substitute \[\left( h,k \right)\] in place of \[\left( {{x}_{2}},{{y}_{2}} \right)\]and \[(1,0)\] in place of \[\left( {{x}_{1}},{{y}_{1}} \right)\] in the distance formula .

So, the distance of \[\left( h,k \right)\] from \[(1,0)\] is given as

\[{{d}_{1}}=\sqrt{{{\left( h-1 \right)}^{2}}+{{\left( k-0 \right)}^{2}}}\]

\[\Rightarrow {{d}_{1}}=\sqrt{{{\left( h-1 \right)}^{2}}+{{k}^{2}}}.......(i)\]

Also , to find the distance of \[\left( h,k \right)\]from \[(-1,0)\], we will substitute \[\left( h,k \right)\] in place of \[\left( {{x}_{2}},{{y}_{2}} \right)\] and \[(-1,0)\] in place of \[\left( {{x}_{1}},{{y}_{1}} \right)\] in the distance formula .

So, the distance of \[\left( h,k \right)\] from \[(-1,0)\]is given as

\[{{d}_{2}}=\sqrt{{{\left( h+1 \right)}^{2}}+{{\left( k-0 \right)}^{2}}}\]

\[\Rightarrow {{d}_{2}}=\sqrt{{{\left( h+1 \right)}^{2}}+{{k}^{2}}}........(ii)\]

Now, in the question it is given that the ratio of distance of vertex from \[(1,0)\] to the distance of vertex from \[(-1,0)\] is equal to \[\dfrac{1}{3}\].

So , \[\dfrac{{{d}_{1}}}{{{d}_{2}}}=\dfrac{1}{3}\]

\[\Rightarrow \dfrac{\sqrt{{{\left( h-1 \right)}^{2}}+{{k}^{2}}}}{\sqrt{{{\left( h+1 \right)}^{2}}+{{k}^{2}}}}=\dfrac{1}{3}\]

Now , we will square both sides to remove the square root sign .

On squaring both sides , we get

\[\begin{align}

& \dfrac{{{\left( h-1 \right)}^{2}}+{{k}^{2}}}{{{\left( h+1 \right)}^{2}}+{{k}^{2}}}=\dfrac{1}{9} \\

& \Rightarrow 9{{\left( h-1 \right)}^{2}}+9{{k}^{2}}={{\left( h+1 \right)}^{2}}+{{k}^{2}} \\

& \Rightarrow 9\left( {{h}^{2}}+1-2h+{{k}^{2}} \right)={{h}^{2}}+1+2h+{{k}^{2}} \\

& \Rightarrow 9{{h}^{2}}+9-18h+9{{k}^{2}}={{h}^{2}}+2h+1+{{k}^{2}} \\

& \Rightarrow 8{{h}^{2}}+8{{k}^{2}}-20h+8=0 \\

\end{align}\]

Taking \[8\] common from the LHS , we get

\[8\left[ {{h}^{2}}+{{k}^{2}}-\dfrac{5}{2}h+1 \right]=0\]

\[\Rightarrow {{h}^{2}}+{{k}^{2}}-\dfrac{5}{2}h+1=0\]

Now , we can write $1$ as $\dfrac{16}{16}$ , which can further be written as $\dfrac{25}{16}-\dfrac{9}{16}$.

So, \[{{h}^{2}}+{{k}^{2}}-\dfrac{5}{2}h+1=0\] can be written as \[{{h}^{2}}-\left( 2\times \dfrac{5}{4}\times h \right)+{{\left( \dfrac{5}{4} \right)}^{2}}-\dfrac{9}{16}+{{k}^{2}}=0\]

\[\begin{align}

& \Rightarrow {{\left( h-\dfrac{5}{4} \right)}^{2}}+{{k}^{2}}-\dfrac{9}{16}=0 \\

& \Rightarrow {{\left( h-\dfrac{5}{4} \right)}^{2}}+{{k}^{2}}=\dfrac{9}{16} \\

\end{align}\]

Now, the locus of \[(h,k)\] is given by replacing \[(h,k)\]by \[(x,y)\].

So, the locus of the vertex of the triangle \[\Delta ABC\] is given as,

\[{{\left( x-\dfrac{5}{4} \right)}^{2}}+{{y}^{2}}={{\left( \dfrac{3}{4} \right)}^{2}}.....(iii)\]

Now, we know the equation of the circle with centre at \[(a,b)\] and radius \[r\] is given as

\[{{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}......(iv)\]

Comparing equation \[(iii)\]and\[(iv)\], we can conclude that the locus of vertex of the triangle \[\Delta ABC\] is a circle with centre at \[\left( \dfrac{5}{4},0 \right)\] and radius \[\dfrac{3}{4}\]units.

Now , we know , the circle on which all the three vertices of a triangle lie is known as the circumcircle and its centre is known as the circumcentre .

So , circumcentre of \[\Delta ABC\] is \[\left( \dfrac{5}{4},0 \right)\]

Option (b) \[\left( \dfrac{5}{4},0 \right)\] is correct.

Note: The distance between two points \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\] is given as \[d=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}\] and not \[d=\sqrt{{{({{x}_{1}}+{{x}_{2}})}^{2}}+{{({{y}_{1}}+{{y}_{2}})}^{2}}}\]. It is a very common mistake made by students.

The given triangle is \[\Delta ABC\].

We will consider \[\left( h,k \right)\] to be coordinates of one of the vertices .

First , we need to find the distance of \[\left( h,k \right)\] from \[(1,0)\]and \[(-1,0)\].

We know that the distance between the two points \[\left( {{x}_{1}},{{y}_{1}} \right)\] and \[\left( {{x}_{2}},{{y}_{2}} \right)\] given as \[d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\]

So , to find the distance of \[\left( h,k \right)\] from \[(1,0)\], we will substitute \[\left( h,k \right)\] in place of \[\left( {{x}_{2}},{{y}_{2}} \right)\]and \[(1,0)\] in place of \[\left( {{x}_{1}},{{y}_{1}} \right)\] in the distance formula .

So, the distance of \[\left( h,k \right)\] from \[(1,0)\] is given as

\[{{d}_{1}}=\sqrt{{{\left( h-1 \right)}^{2}}+{{\left( k-0 \right)}^{2}}}\]

\[\Rightarrow {{d}_{1}}=\sqrt{{{\left( h-1 \right)}^{2}}+{{k}^{2}}}.......(i)\]

Also , to find the distance of \[\left( h,k \right)\]from \[(-1,0)\], we will substitute \[\left( h,k \right)\] in place of \[\left( {{x}_{2}},{{y}_{2}} \right)\] and \[(-1,0)\] in place of \[\left( {{x}_{1}},{{y}_{1}} \right)\] in the distance formula .

So, the distance of \[\left( h,k \right)\] from \[(-1,0)\]is given as

\[{{d}_{2}}=\sqrt{{{\left( h+1 \right)}^{2}}+{{\left( k-0 \right)}^{2}}}\]

\[\Rightarrow {{d}_{2}}=\sqrt{{{\left( h+1 \right)}^{2}}+{{k}^{2}}}........(ii)\]

Now, in the question it is given that the ratio of distance of vertex from \[(1,0)\] to the distance of vertex from \[(-1,0)\] is equal to \[\dfrac{1}{3}\].

So , \[\dfrac{{{d}_{1}}}{{{d}_{2}}}=\dfrac{1}{3}\]

\[\Rightarrow \dfrac{\sqrt{{{\left( h-1 \right)}^{2}}+{{k}^{2}}}}{\sqrt{{{\left( h+1 \right)}^{2}}+{{k}^{2}}}}=\dfrac{1}{3}\]

Now , we will square both sides to remove the square root sign .

On squaring both sides , we get

\[\begin{align}

& \dfrac{{{\left( h-1 \right)}^{2}}+{{k}^{2}}}{{{\left( h+1 \right)}^{2}}+{{k}^{2}}}=\dfrac{1}{9} \\

& \Rightarrow 9{{\left( h-1 \right)}^{2}}+9{{k}^{2}}={{\left( h+1 \right)}^{2}}+{{k}^{2}} \\

& \Rightarrow 9\left( {{h}^{2}}+1-2h+{{k}^{2}} \right)={{h}^{2}}+1+2h+{{k}^{2}} \\

& \Rightarrow 9{{h}^{2}}+9-18h+9{{k}^{2}}={{h}^{2}}+2h+1+{{k}^{2}} \\

& \Rightarrow 8{{h}^{2}}+8{{k}^{2}}-20h+8=0 \\

\end{align}\]

Taking \[8\] common from the LHS , we get

\[8\left[ {{h}^{2}}+{{k}^{2}}-\dfrac{5}{2}h+1 \right]=0\]

\[\Rightarrow {{h}^{2}}+{{k}^{2}}-\dfrac{5}{2}h+1=0\]

Now , we can write $1$ as $\dfrac{16}{16}$ , which can further be written as $\dfrac{25}{16}-\dfrac{9}{16}$.

So, \[{{h}^{2}}+{{k}^{2}}-\dfrac{5}{2}h+1=0\] can be written as \[{{h}^{2}}-\left( 2\times \dfrac{5}{4}\times h \right)+{{\left( \dfrac{5}{4} \right)}^{2}}-\dfrac{9}{16}+{{k}^{2}}=0\]

\[\begin{align}

& \Rightarrow {{\left( h-\dfrac{5}{4} \right)}^{2}}+{{k}^{2}}-\dfrac{9}{16}=0 \\

& \Rightarrow {{\left( h-\dfrac{5}{4} \right)}^{2}}+{{k}^{2}}=\dfrac{9}{16} \\

\end{align}\]

Now, the locus of \[(h,k)\] is given by replacing \[(h,k)\]by \[(x,y)\].

So, the locus of the vertex of the triangle \[\Delta ABC\] is given as,

\[{{\left( x-\dfrac{5}{4} \right)}^{2}}+{{y}^{2}}={{\left( \dfrac{3}{4} \right)}^{2}}.....(iii)\]

Now, we know the equation of the circle with centre at \[(a,b)\] and radius \[r\] is given as

\[{{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}......(iv)\]

Comparing equation \[(iii)\]and\[(iv)\], we can conclude that the locus of vertex of the triangle \[\Delta ABC\] is a circle with centre at \[\left( \dfrac{5}{4},0 \right)\] and radius \[\dfrac{3}{4}\]units.

Now , we know , the circle on which all the three vertices of a triangle lie is known as the circumcircle and its centre is known as the circumcentre .

So , circumcentre of \[\Delta ABC\] is \[\left( \dfrac{5}{4},0 \right)\]

Option (b) \[\left( \dfrac{5}{4},0 \right)\] is correct.

Note: The distance between two points \[({{x}_{1}},{{y}_{1}})\] and \[({{x}_{2}},{{y}_{2}})\] is given as \[d=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}\] and not \[d=\sqrt{{{({{x}_{1}}+{{x}_{2}})}^{2}}+{{({{y}_{1}}+{{y}_{2}})}^{2}}}\]. It is a very common mistake made by students.

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