Question

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

• A. $x=0,y=0$
• B. $x=1,y=0$
• C. $(h^2–r^2)x–2rhy=0,y=0$
• D. $(h^2–r^2)x–2rhy=0,x=0$

Answer 1

The correct option is D

$(h^2–r^2)x–2rhy=0,x=0$

Explanation for correct answer:

Finding the equation of tangents drawn from the origin :

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

As we know, the general equation circle, is ${(x-a)}^2+{(y-b)}^2=r^2$

Converting the given equation into this form.

Now, adding $r^2$ on both sides we get,

$$\begin{gathered} \Rightarrow x^2+y^2-2rx-2hy+h^2+r^2=r^2 \\ \Rightarrow \left(x^2-2rx+r^2\right)+\left(y^2-2hy+h^2\right)=r^2 \\ \Rightarrow {(x-r)}^2+{(y-h)}^2=r^2 \end{gathered}$$

Here $(r,h)$is center and $r$ is the radius.

The tangent is touching at $x=0$ because radius and $x$ coordinate of the center are equal

Therefore, $x=0$ is a tangent.

So, the given circle has $y$-axis as one tangent.

Let $y=mx$ be another tangent with equation $[y-mx=0]$

Distance from $(r,h)=r$.

Length of the perpendicular from origin $=$ Radius

$\Rightarrow \left|\frac{h-mr}{\sqrt{1+m^2}}\right|=r$

$$\begin{gathered} \Rightarrow m^2{r}^2+h^2+2mrh=r^2\left(m^2+1\right) \\ \Rightarrow m^2{r}^2+h^2+2mrh=m^2{r}^2+r^2 \\ \Rightarrow 2mrh=r^2-h^2 \\ \Rightarrow m=\left|\frac{r^2-h^2}{2rh}\right| \end{gathered}$$

Hence, the equation of tangent, $y=\left|\frac{r^2-h^2}{2rh}\right|x$

​Another tangent will be $(h^2-r^2)x-2rhy=0$

Hence, the correct answer is option (D).