JEE MAIN - Mathematics (2009 - No. 14)
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 :
$$\left( {{5 \over 4},0} \right)$$
$$\left( {{5 \over 2},0} \right)$$
$$\left( {{5 \over 3},0} \right)$$
$$\left( {0,0} \right)$$
Explanation
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 $$\left( 1 \right) = \left( {{5 \over 4},0} \right)$$
$$P\left( {1,0} \right),Q\left( { - 1,0} \right)$$
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 $$\left( 1 \right) = \left( {{5 \over 4},0} \right)$$
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