Three distinct points A, B and C are given in the 2 -dimensional coordinates plane such th…

Question

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

Accepted Answer

Given that

$$P\left( {1,0} \right),Q\left( { - 1,0} \right)$$

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

$$ \Rightarrow 3AP = AQ$$

$$\,\,\,\,\,\,$$ Let $A = (x,y)$ then $$3AP = AQ \Rightarrow 9A{P^2} = A{Q^2}$$

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

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

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

$$ \Rightarrow {x^2} + {y^2} - {5 \over 3}x + 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$$\therefore$$ A lies on  the circle given by  eq. $(1).$ As $B$ and $C$

also follow the same condition, - they must lie on the same circle.

$$\therefore$$ Center of circumcircle of $$\Delta ABC$$

$=$ Center of circle given  by  $$\left( 1 \right) = \left( {{5 \over 4},0} \right)$$

Solutions · 1