If the line `(x/a)+(y/b)=1` moves in such a way that `(1/(a^2))+(1/(b^2))=(1/(c^2))` , where `c` is a constant, prove that the foot of the perpendicular from the origin on the straight line describes the circle `x^2+y^2=c^2dot`

The line (x)/(a) + (y)/( b) =1 moves in such a way that (1) /( a^2) + (1)/( b^2) = (1)/( c^2) , where c is a constant. The locus of the foot of the perpendicular from the origin on the given line is x^2 + y^2 = c^2

The variable coefficients a, b in the equation of the straight line (x)/(a) + (y)/(b) = 1 are connected by the relation (1)/(a^(2)) + (1)/(b^(2)) = (1)/(c^(2)) where c is a fixed constant. Show that the locus of the foot of the perpendicular from the origin upon the line is a circle. Find the equation of the circle.

if (x)/(a)+(y)/(b)=1 is a variable line where (1)/(a^(2))+(1)/(b^(2))=(1)/(c^(2))(c is constant ) then the locus of foot of the perpendicular drawn from origin

if x/a+y/b=1 is a variable line where 1/a^2+1/b^2=1/c^2 ( c is constant ) then the locus of foot of the perpendicular drawn from origin

If p is the perpendicular distance from the origin to the straight line x/a+y/b=1 , then 1/(a^(2))+1/(b^(2))=

if P is the length of perpendicular from origin to the line x/a+y/b=1 then prove that 1/(a^2)+1/(b^2)=1/(p^2)