Question

The equation of the tangents drawn from the origin to the circle ${{x}^{2}}+{{y}^{2}}-2rx-2hy+{{h}^{2}}=0$ are $$$$
A.x=0$$$$$ B.y=0 C.$\left( {{h}^{2}}-{{r}^{2}} \right)y-2rhx=0
D. $\left( {{h}^{2}}-{{r}^{2}} \right)y+2rhx=0$$$$$

Answer 1

We use completing square method to convert the given equation of circle ${{x}^{2}}+{{y}^{2}}-2rx-2hy+{{h}^{2}}=0$ with the standard equation of circle in centre-radius from ${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$ to get coordinates of centre $\left( a,b \right)$ and radius $r$. We find that the $y-$ axis is tangent to the circle and we find the slope of $m$ of the other tangent $y-mx=0$ equating the radius with distance from centre to the tangent. $$$$

Complete step by step answer:
We know that the standard equation of a circle with centre $\left( a,b \right)$ and radius $r$ is given by ;
${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{r}^{2}}$
We are given the following equation of circle in the question
${{x}^{2}}+{{y}^{2}}-2rx-2hy+{{h}^{2}}=0$
Let us use completing square method in the above equation to have;

& {{x}^{2}}-2rx+{{r}^{2}}+{{y}^{2}}-2hy+{{h}^{2}}={{r}^{2}} \\\ & \Rightarrow {{\left( x-r \right)}^{2}}+{{\left( y-h \right)}^{2}}={{r}^{2}} \\\ \end{aligned}$$ We compare the above equation with the standard equation of the circle and find the centre as $\left( r,h \right)$ and radius as $r$. Since the $x-$ coordinate of the centre is equal to the radius then the $y-$ axis or the equation $x=0$ touches the circle. Let the other tangent of the circle has slope $m$.Since it passes through the origin we can write its equation as; $$\begin{aligned} & y=mx \\\ & \Rightarrow y-mx=0.......\left( 2 \right) \\\ \end{aligned}$$ Since radius of circle is equal to the distance from the centre of the circle $\left( r,h \right)$to the tangent line $y-mx=0$ , we have; $$r=\left| \dfrac{r+\left( -m \right)h}{\sqrt{1+{{\left( -m \right)}^{2}}}} \right|=\left| \dfrac{r-mh}{\sqrt{1+{{m}^{2}}}} \right|$$ We square both sides and cross multiply to have; $$\begin{aligned} & \Rightarrow {{r}^{2}}\left( 1+{{m}^{2}} \right)={{\left( r-mh \right)}^{2}} \\\ & \Rightarrow {{r}^{2}}+{{r}^{2}}{{m}^{2}}={{r}^{2}}+{{m}^{2}}{{h}^{2}}-2rmh \\\ & \Rightarrow {{r}^{2}}{{m}^{2}}-{{m}^{2}}{{h}^{2}}+2rmh=0 \\\ & \Rightarrow m\left( {{r}^{2}}m-m{{h}^{2}}+2rh \right)=0 \\\ \end{aligned}$$ So we obtain the roots as $$\begin{aligned} & \Rightarrow m=0\text{ or }{{r}^{2}}m-m{{h}^{2}}+2rh=0 \\\ & \Rightarrow m=0\text{ or m}\left( {{r}^{2}}-{{h}^{2}} \right)=-2rh \\\ & \Rightarrow m=0\text{ or }m=\dfrac{2rh}{{{h}^{2}}-{{r}^{2}}} \\\ \end{aligned}$$ So the equation of the other tangent is $$\begin{aligned} & y=\dfrac{2rh}{{{h}^{2}}-{{r}^{2}}}x \\\ & \Rightarrow y\left( {{h}^{2}}-{{r}^{2}} \right)=2rhx \\\ & \Rightarrow \left( {{h}^{2}}-{{r}^{2}} \right)y-2rhx=0 \\\ \end{aligned}$$ So the equations of tangents are $x=0,\left( {{h}^{2}}-{{r}^{2}} \right)y-2rhx=0$.  **So, the correct answer is “Option c”.** **Note:** We note that the distance between a point $\left( {{x}_{1}},{{y}_{1}} \right)$ from a line $ax+by+c=0$ is given by $d=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$. The general equation of line passing with slope $m$ and $y-$intercept $c$ is given by $y=mx+c$ and all lines passing through origin are given by $y=mx$. The condition of perpendicular between the tangents will be $r=\pm h$.