A point moves so that the sum of the squares of its distances from the angular points of a triangle is constant; prove that its locus is a circle.

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Solution

Let given $△$ is $A B C$ with angular points $A (x_{1}, y_{1}), B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$

Let the moving point be $P (h, k)$

Given ${P A}^{2} + {P B}^{2} + {P C}^{2} = c$

${(h - x_{1})}^{2} + {(k - y_{1})}^{2} + {(h - x_{2})}^{2} + {(k - y_{2})}^{2} + {(h - x_{3})}^{2} + {(k - y_{3})}^{2} = c$
$h^{2} + x_{1}^{2} - 2 a h x_{1} + k^{2} + y_{1}^{2} - 2 a h y_{1} + h^{2} + x_{2}^{2} - 2 a h x_{2} + k^{2} + y_{2}^{2} - 2 a h y_{2} + h^{2} +$
$x_{3}^{2} - 2 a h x_{3} + k^{2} + y_{3}^{2} - 2 a h y_{3} = c$
$3 h^{2} + 3 k^{2} - 2 a h {(x}_{1} + x_{2} + x_{3}) - 2 a k {(y}_{1} + y_{2} + y_{3}) + x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + y_{1}^{2} + y_{2}^{2} + y_{3}^{2} = c$
$h^{2} + k^{2} - \frac{2 a h}{3} {(x}_{1} + x_{2} + x_{3}) - \frac{2 a k}{3} {(y}_{1} + y_{2} + y_{3}) + \frac{x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + y_{1}^{2} + y_{2}^{2} + y_{3}^{2} - c}{3} = 0$

Generalising the equation we get

$x^{2} + y^{2} - \frac{2 a x}{3} {(x}_{1} + x_{2} + x_{3}) - \frac{2 a y}{3} {(y}_{1} + y_{2} + y_{3}) + \frac{x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + y_{1}^{2} + y_{2}^{2} + y_{3}^{2} - c}{3} = 0$

Clearly the equation represents a circle with centre $(\frac{a {(x}_{1} + x_{2} + x_{3})}{3}, \frac{a {(y}_{1} + y_{2} + y_{3})}{3})$

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