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the locus of the foot of the perpendicular draw from origin to a variable line passing through fixed point (2,3) is a circle whose diameter is


Aditya Shelote 1st Dec, 2020
Answer (1)
Shruti Smaranika 8th Dec, 2020

Hello aspirant,

The answer will be-

Assume slope of line be m

(2,3) are fixed points

Equation of line:-

y-3=m*(x-2)

y-3=mx-2m

mx-y+3-2m=0

Foot of perpendicular from origin (0,0) is-

(2-0)/m=(y-0)/(-1)= -(0+3-2m)/{0.5^(1+m^2)}

m= -x/y

Now, mx-y+3-2m=0

(-x/y)x-y+3-2(-x/y)=0

x^2+y^2-2^x-3^y=0

Centre=(1, 3/2)

Radius= 0.5^(1^2+(3/2)^2-0) = √13/2

Diameter = 2*Radius = 2* (√13/2) = √13

Hope, it helps you.

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