Question

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

• A. $x=0,y=0$
• B. $(h^2-r^2)x-2rhy=0,x=0$
• C. $y=0,x=4$
• D. $(h^2-r^2)x+2rhy=0,x=0$

Answer 1

Answer: (B)

$(h^2-r^2)x-2rhy=0,x=0$

REF.Image.

Solution-

$x^2+y^2-2rx-2hy+h^2=0$

$x^2+y^2-2rx-2hy+h^2+r^2=r^2$

$(x-r{)}^2+(y-h{)}^2=r^2$

(r,h) is centre, r is radius

It is touching x = 0 because radius $△x$ coordinate of

centre are equal

x =0 is a tangent

Let y = mx be another tangent [ y-mx = 0]

1 distance from (r,h) = r.

$\left|\frac{r-mh}{\sqrt{1+m^2}}\right|=r.$ $m=0,\frac{h^2-r^2}{2rh}$

Another tangent will be $(h^2-r^2)x-2rhy=0$ B is correct.