Question

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

• A. $y=0$
• B. $x-y=0$
• C. $(h^2-r^2)x-2rhy=0$
• D. None of these

Answer 1

The correct option is C $(h^2-r^2)x-2rhy=0$

Let the tangent from the origin touches the circle $x^2+y^2-2rx-2hy+h^2=0$.......(1) at $P(x_1,y_1)$.

Equation (1) can be written as $(x-r{)}^2+(y-h{)}^2=r^2$.

So, the centre of the circle (1) is $(r,h)$ and radius $=r$.

Again the length of the tangent from the origin is $h$ and the distance between the point $P$ to the origin $=\sqrt{x_1^2+y_1^2}$.

$\sqrt{x_1^2+y_1^2}=h$

or, $x_1^2+y_1^2=h^2$......(2).

Again equation of tangent to the circle (1) at $(x_1,y_1)$ is

$x{x}_1+y{y}_1-r(x+x_1)-h(y+y_1)+h^2=0$ which passes through $(0,0)$ then

or, $-r{x}_1-h{y}_1+h^2=0$

or, $r{x}_1+h{y}_1=h^2$......(3).

Now solving from (2) and (3) we get,

$x_1=\frac{2h{r}^2}{h^2+r^2},y_1=\frac{h(h^2-r^2)}{h^2+r^2}$.

The equation of tangent passing through the origin is

$y=\frac{y_1}{x_1}x$

or, $y=\frac{(h^2-r^2)x}{2rh}$ [Using the values of $x_1$ and $y_1$]

or, $(h^2-r^2)x-2hry=0$.