The equation of tangents drawn from the origin to the circle`x^2+y^2-2rx-2hy+h^2=0`

Options

A. `x=0`
B. `y=0`
C. `(h^(2)-r^(2))x-2rhy=0`
D. `(h^(2)-r^(2))x+2hy=0`

Answer

1,3

Explanation

The equation of any line through the origin (0,0) is
`y=mx` (1)
If line (i) is tangent to the circle `x^(2)+y^(2)-2rx-2hy+h^(2)=0`, then the length of perpendicular from center (r,h) on (i) is equal to the radius of the circle, i.e.,
`(|mr-h|)/(sqrt(m^(2)+1))=sqrt(r^(2)+h^(2)-h^(2))`
`(mr-h)^(2)=(m^(2)+1)r^(2)`
`0.m^(2)+(2hr)m+(r^(2)-h^(2))=0`
`m=oo,(h^(2)-r^(2))/(2hr)`