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(A) Let the point $P$ be $(x, y)$.According to the problem,the distance from $(5, 0)$ is thrice the distance from $(-5, 0)$:$\sqrt{(x-5)^{2} + y^{2}}

Vedclass · Accepted Answer

$56$

Vedclass · Answer

(A) Let the point $P$ be $(x, y)$.According to the problem,the distance from $(5, 0)$ is thrice the distance from $(-5, 0)$:$\sqrt{(x-5)^{2} + y^{2}} = 3\sqrt{(x+5)^{2} + y^{2}}$Squaring both sides:$(x-5)^{2} + y^{2} = 9((x+5)^{2} + y^{2})$$x^{2} - 10x + 25 + y^{2} = 9(x^{2} + 10x + 25 + y^{2})$$x^{2} - 10x + 25 + y^{2} = 9x^{2} + 90x + 225 + 9y^{2}$$8x^{2} + 8y^{2} + 100x + 200 = 0$Dividing by $8$:$x^{2} + y^{2} + 12.5x + 25 = 0$Comparing with the standard circle equation $x^{2} + y^{2} + 2gx + 2fy + c = 0$,we have $g = 6.25$,$f = 0$,and $c = 25$.The radius $r$ is given by $r = \sqrt{g^{2} + f^{2} - c} = \sqrt{(6.25)^{2} - 25} = \sqrt{39.0625 - 25} = \sqrt{14.0625}$.Thus,$r^{2} = 14.0625$.Therefore,$4r^{2} = 4 	imes 14.0625 = 56.25$.

Let a point $P$ be such that its distance from the point $(5, 0)$ is thrice the distance of $P$ from the point $(-5, 0)$. If the locus of the point $P$ is a circle of radius $r$,then $4r^{2}$ is equal to ...... .

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