# The locus of the foot of the perpendicular from the origin to the line which always passes through a fixed point (h,k) is

Arun Kumar IIT Delhi
9 years ago
Hello Student,

y = mx+c
Foot of perpendicular from (0, 0) to the line be (a, b)
$a = \frac{-mc}{1+m^{2}}$
$b = \frac{c}{1+m^{2}}$
$m = \frac{-a}{b}$
Since line passes through (h, k):
$k = mh +c$
$c = k-mh$
$b = \frac{k-mh}{1+m^{2}}$
$b = \frac{k-(\frac{-a}{b})h}{1+(\frac{-a}{b})^{2}}$
$a^{2}+b^{2}-ah-bk=0$
$x^{2}+y^{2}-ax-by=0$
$a\rightarrow x, b\rightarrow y$
=>Equation of circle.

Thanks & Regards
Arun Kumar
Btech, IIT Delhi
ram
11 Points
6 years ago
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11 Points
6 years ago
Equation of any line through point (h,k) is y-k=m(x-h)
Equation of the line perpendicular to the line through origin is y=(-1/m)x
To get the locus of the intersection of the above two lines i,e foot of the perpendicular to first line, eleminate ‘m’ from the above two equations.
Multiplying the two equations, y(y-k)=m(-1/m)(x-h)x which is same as x^2 + y^2 – hx – ky = 0