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 circumcentre of the triangle ABC is at the point:

• A. $(\frac{5}{2},0)$
• B. $(\frac{5}{3},0)$
• C. $(0,0)$
• D. $(\frac{5}{8},0)$

Answer 1

Answer: (D)

$(\frac{5}{8},0)$

Let (h,k) be the locus of points A, B & C

using given condition

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

$9(h-1{)}^2+9K^2=(h+1{)}^2+K^2$

or,

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

$h^2+K^2-\frac{5h}{4}+1=0$

this is a circle and it will be same circle which circumscribe $\Delta ABC$ as it contains all three pts on it.

$\therefore$ center of circumcircle $\equiv (\frac{5}{8},0)$