Question

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

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

Answer 1

Answer: (A), (C)

The correct options are
A $x=0$
C $(h^2-r^2)x-2rhy=0$

The equation of any line through the origin (0, 0) is y = mx . . .(i)
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.,
$\frac{|mr-h|}{\sqrt{m^2+1}}=\sqrt{r^2+h^2-h^2}(mr-h{)}^2=(m^2+1)r^{2}0.m^2+(2hr)m+(r^2-h^2)=0m=\infty ,\frac{h^2-r^2}{2hr}$
Substituting these values in (i), we get the tangents as
$x=0and(h^2-r^2)x-2rhy=0$