Question

Three distinct points $A,B$ and $C$ are 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 $\frac{1}{3}$ . Circumcenter of the triangle $ABC$ is at the point $(\frac{a}{b},0),a,b\in N$ and coprime. Then $a+b$ is equal to

• A. 9.0
• B. 09
• C. 9
• D. 9.00

Answer 1

Answer: (A), (B), (C), (D)

Let the point be $(h,k)$
$\therefore \frac{\sqrt{(h-1{)}^2+k^2}}{\sqrt{(h+1{)}^2+k^2}}=\frac{1}{3}$
$\Rightarrow \left[\right(h-1{)}^2+k^2]9=(h+1{)}^2+k^2$
$\Rightarrow 9(h^2+k^2+1-2h)=h^2+k^2+1+2h$
$\Rightarrow 8h^2+8k^2-20h+8=0$
$\Rightarrow h^2+k^2-\frac{5}{2}h+1=0$
$\Rightarrow h^2-\frac{5}{2}h+\frac{25}{16}+k^2=\frac{25}{16}-1$
$\Rightarrow {(h-\frac{5}{4})}^2+k^2=\frac{9}{16}$

Hence, locus of point $A,B,C:$
${(x-\frac{5}{4})}^2+y^2={\left(\frac{3}{4}\right)}^2$
Thus, the locus will be of circle with centre at $(\frac{5}{4},0)$
$\therefore$ The circumcenter of any triangle formed by A, B, C will lie on the centre of circle.
Hence, $a+b=5+4=9$