Questions & Answers

Question

Answers

A.$\left( {\dfrac{5}{4},0} \right)$

B.$\left( {\dfrac{5}{2},0} \right)$

C.$\left( {\dfrac{5}{3},0} \right)$

D.$\left( {0,0} \right)$

Answer
Verified

Let the coordinates of the point be $\left( {h,k} \right)$ whose distance from the point $\left( {1,0} \right)$ to the distance from the point $\left( { - 1,0} \right)$ is equal to $\dfrac{1}{3}$.

We will first find the distance from $\left( {h,k} \right)$ to $\left( {1,0} \right)$

If $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ are two points, then the distance between the points is given by $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $

Then, the distance from $\left( {h,k} \right)$ to $\left( {1,0} \right)$ will be $\sqrt {{{\left( {h - 1} \right)}^2} + {k^2}} $

Similarly, the distance from $\left( {h,k} \right)$ to $\left( { - 1,0} \right)$ will be $\sqrt {{{\left( {h + 1} \right)}^2} + {k^2}} $

We are given that the ratio of both the distances is $\dfrac{1}{3}$

Hence, $\dfrac{{\sqrt {{{\left( {h - 1} \right)}^2} + {k^2}} }}{{\sqrt {{{\left( {h + 1} \right)}^2} + {k^2}} }} = \dfrac{1}{3}$

On squaring both sides, we will get,

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

Cross-multiply and simplify

$

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

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

$

On solving the brackets, we get

$

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} + 1 + 2h + {k^2} \\

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

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

$

We have to find the circumcentre of the triangle ABC.

Circumference is the centre of the circle which passes through each point on the circle.

The equation ${h^2} + {k^2} - \dfrac{5}{2}h + 1 = 0$ represents the equation of the circle.

Compare it with the standard equation of the circle to find the circumcentre.

The standard equation of the circle is ${x^2} + {y^2} + 2gx + 2fy + c = 0$, whose centre is $\left( { - g, - f} \right)$

Here, $2g = - \dfrac{5}{2}$ and $2f = 0$

Hence, $g = - \dfrac{5}{4}$ and $f = 0$

Therefore, the coordinates of circumcentre is $\left( {\dfrac{5}{4},0} \right)$