Question

Three distinct points A, B and C are given in the $2-$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 $\frac{1}{3}$. Then the circumcenter of the triangle ABC is at the point:

• A. $(\frac{5}{4},0)$
• B. $(\frac{5}{2},0)$
• C. $(\frac{5}{3},0)$
• D. $0,0$

Answer 1

The correct option is A $(\frac{5}{4},0)$

Let the point be h,k

$\therefore \frac{\sqrt{{(h-1)}^2+k^2}}{\sqrt{{(h+1)}^2+k^2}}=\frac{1}{3}$

$\therefore [{(h-1)}^2+k^2]9={(h+1)}^2+k^2$

$9[h^2+k^2+1-2h]=h^2+k^2+1+2h$

$9h^2+9k^2+9-18h=h^2+k^2+1+2h$

$8h^2+8k^2-20h+8=0$

$8[h^2+k^2-\frac{5}{2}h+1]=0$

$h^2-\frac{5}{2}h+\frac{25}{16}+k^2=\frac{25}{16}-1$

${(h-\frac{5}{4})}^2+k^2=\frac{9}{16}$

Locus of the points A,B and C:

${(x-\frac{5}{4})}^2+y^2={\left(\frac{3}{4}\right)}^2$

Thus, the locus will be of a circle with center at $(\frac{5}{4},0)$ and radius $\frac{3}{4}$ units

The circumcenter of any triangle formed by A,B and C will lie on the center of the circle.