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  • 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 <img al

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 \mathrm{4 r^2}  is equal to

Option: 1

60.15


Option: 2

25.56


Option: 3

52.65


Option: 4

56.25


Answers (1)

best_answer

\mathrm{\sqrt{(h-5)^2+k^2}=3 \sqrt{(h+5)^2+k^2} \\ }

\mathrm{ \Rightarrow h^2+k^2+25-10 h=9\left(h^2+k^2+25+10 h\right) \\ }

\mathrm{ \Rightarrow 8 h^2+8 k^2+100 h+200=0 \\ }

\mathrm{ \Rightarrow h^2+k^2+\frac{25}{2} h+25=0 }

Thus, the locus is\mathrm{ x^2+y^2+\frac{25}{2} x+25=0 },

which is a circle whose centre is \mathrm{ \left(-\frac{25}{4}, 0\right) } and radius, \mathrm{ r=\sqrt{\left(\frac{-25}{4}\right)^2+(0)^2-25} }

\mathrm{\Rightarrow 4 r^2=\frac{(25)^2-400}{4}=\frac{225}{4}=56.25 }.

Posted by

himanshu.meshram

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