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. $(\frac{5}{4},0)$

Answer 1

Answer: (D)

$(\frac{5}{4},0)$

Let $A(x,y)$

$P=(1,0)$ , $Q=(-1,0)$

$\frac{AP}{AQ}=\frac{BP}{BQ}=\frac{CP}{CQ}=\frac{1}{3}$

$3AP=AQ$

$9A{P}^2=A{Q}^2$

$9(x-1{)}^2+9y^2=(x+1{)}^2+y^2$

$9x^2-18x+9+9y^2=x^2+2x+1+y^2$

$8x^2-20x+8y^2+8=0$

$x^2+y^2-\frac{5}{2}x+1=0$

Therefore A lies on circle

Similarly B and C lie on circle also

Circumcentre of $△ABC$ = Centre of circle

$=(\frac{5}{4},0)$ Ans