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(A) Let $P = (1, 0)$ and $Q = (-1, 0)$. Let a point $X = (x, y)$ satisfy the condition $\frac{XP}{XQ} = \frac{1}{2}$.This implies $2XP = XQ$,or $4XP^2

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$\left( \frac{5}{3}, 0 \right)$

Vedclass · Answer

(A) Let $P = (1, 0)$ and $Q = (-1, 0)$. Let a point $X = (x, y)$ satisfy the condition $\frac{XP}{XQ} = \frac{1}{2}$.This implies $2XP = XQ$,or $4XP^2 = XQ^2$.Substituting the coordinates: $4((x - 1)^2 + y^2) = (x + 1)^2 + y^2$.Expanding this: $4(x^2 - 2x + 1 + y^2) = x^2 + 2x + 1 + y^2$.$4x^2 - 8x + 4 + 4y^2 = x^2 + 2x + 1 + y^2$.$3x^2 + 3y^2 - 10x + 3 = 0$.Dividing by $3$: $x^2 + y^2 - \frac{10}{3}x + 1 = 0$.This is the equation of a circle. Since points $A, B, C$ satisfy this condition,they lie on this circle.The circumcentre of $\Delta ABC$ is the center of this circle.Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we have $2g = -\frac{10}{3} \Rightarrow g = -\frac{5}{3}$ and $f = 0$.The center is $(-g, -f) = \left( \frac{5}{3}, 0 \right)$.

If three distinct points $A, B, C$ are given in the $2$-dimensional coordinate plane such that the ratio of the distance from $(1, 0)$ to the distance from $(-1, 0)$ is equal to $\frac{1}{2}$,then the circumcentre of the triangle $ABC$ is at the point

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