Question

Three distinct points $\mathrm{A},\mathrm{B}$ and $\mathrm{C}$ given in the two-dimensional coordinate plane such that the ratio of the distance of any one of them from the point $(1,0)$ to the distance from the point $(-1,0)$ is equal to $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$. Then, the circumcenter of the triangle $\mathrm{ABC}$ is at the point

• A. $(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,0)$
• B. $(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.,0)$
• C. $(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.,0)$
• D. $\left(0,0\right)$

Answer 1

The correct option is A

$(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.,0)$

Explanation for the correct answer.

Let $\mathrm{P}(\mathrm{x},\mathrm{y})$ be a general point such that $\frac{\mathrm{PM}}{\mathrm{PN}}=\frac{1}{3}$ where $M(1,0)$ and $N(-1,0)$

Distance Formula $=\sqrt{{({\mathrm{x}}_2-{\mathrm{x}}_1)}^2+{({\mathrm{y}}_2-{\mathrm{y}}_1)}^2}$

$\begin{array}{crcl}\Rightarrow & \frac{\sqrt{{(x-1)}^2+{(y-0)}^2}}{\sqrt{{\left(x+1\right)}^2+{\left(y-0\right)}^2}}& =& \frac{1}{3}\\ \Rightarrow & \frac{\sqrt{{\left(x-1\right)}^2+y^2}}{\sqrt{{\left(x+1\right)}^2+y^2}}& =& \frac{1}{3}\\ \Rightarrow & 3\sqrt{{(x-1)}^2+y^2}& =& \sqrt{{(x+1)}^2+y^2}\\ \Rightarrow & 9\left[x^2+1-2x+y^2\right]& =& x^2+1+2x+y^2\\ \Rightarrow & 9x^2+9-18x+9y^2& =& x^2+1+2x+y^2\\ \Rightarrow & 8x^2+8y^2-20x+8& =& 0\\ \Rightarrow & x^2+y^2-\frac{5}{2}x+1& =& 0\end{array}$

The general equation of the circle ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0$

$2g=\frac{-5}{2},2f=0$

Radius is given by $\left(-g,-f\right)$

Compared with a general equation we get $\left(\frac{5}{4},0\right)$

Hence option A is the correct answer.