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 1:3 then the circumference of the triangle ABC is at the point

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

Answer 1

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

We have,

Let point

$P(x_1,y_1)=(1,0)$

$Q(x_2,y_2)=(-1,0)$

Let the point $A(x,y)$

According to given question,

$\frac{AP}{AQ}=\frac{BP}{BQ}=\frac{CP}{CQ}=\frac{1}{3}......\left(1\right)$

$\frac{AP}{AQ}=\frac{1}{3}$

$\Rightarrow 3AP=AQ$

Squaring both side and we get,

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

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

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

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

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

Therefore point A lies on the circle

Circumference of $\Delta ABC$ =centre of circle $=(\frac{5}{4},0)$

Hence, this is the answer.