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
Answer (1)
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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