Three distinct points A, B and C are given in the 2-dimensional plane such that the ratio of the distance of any of them from the point \[(1,0)\] to the distance from the point \[(-1,0)\] is equal to \[\dfrac{1}{3}\]. Then the circumcentre of triangle ABC is at the point
A) \[(0,0)\]
B) \[\left( \dfrac{5}{4},0 \right)\]
C) \[\left( \dfrac{5}{2},0 \right)\]
D) \[\left( \dfrac{5}{3},0 \right)\]
Answer
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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.
Last updated date: 21st Sep 2023
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