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`

Answer

True

Explanation

Given that, equation of line is
`(x)/(a)+(y)/(b)=1" (i)"`
Equation of line passing through orgin and perpendicular to line (i) is
`(x)/(b)-(y)/(a)=0" (ii)"`
Now, foot of perpendicular is the point of intersection of lines (i) and (ii). To find its locus we have to eliminate the variable a and b.
On squaring and adding Eqs. (i) and (ii), we get
`(x^(2))/(a^(2))+(y^(2))/(b^(2))+(2xy)/(ab)+(x^(2))/(b^(2))+(y^(2))/(a^(2))-(2xy)/(ab)=1`
`rArr x^(2)((1)/(a^(2))+(1)/(b^(2)))+y^(2)((1)/(a^(2))+(1)/(b^(2)))=1`
`rArr (x^(2))/(c^(2))+(y^(2))/(c^(2))=1`
`rArr x^(2)+y^(2)=c^(2)" "[:. (1)/(a^(2))+(1)/(b^(2))=(1)/(c^(2))]`
Hence, the statement is true.