Question

Consider the relation $4{{l}^{2}}-5{{m}^{2}}+6l+1=0$, where $l,m\in R$. Then the line $lx+my+1=0$ touches a fixed circle whose centre and radius are, respectively
A. $\left( 2,0 \right),3$
B. $\left( -3,0 \right),\sqrt{3}$
C. $\left( 3,0 \right),\sqrt{5}$
D. None of these

Answer 1

To solve this question, we should know the relation between the tangent of the circle and the circle itself. The perpendicular distance between the centre of the circle and the tangent of the circle is equal to radius of the circle. The perpendicular distance between a line $ax+by+c=0$ and a point $\left( {{x}_{1}},{{y}_{1}} \right)$ is given by $\text{distance}=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$. Here, we are given an equation of the tangent $lx+my+1=0$ and we are asked to find the centre and radius of the circle using the relation $4{{l}^{2}}-5{{m}^{2}}+6l+1=0$. We have to check option by option by taking the centre coordinates in the option as $\left( {{x}_{1}},{{y}_{1}} \right)$ and the equation $lx+my+1=0$ as $ax+by+c=0$ and the distance as radius in the option. We should simplify and get the relation $4{{l}^{2}}-5{{m}^{2}}+6l+1=0$ and the option which gets this relation is the answer.

Complete step-by-step answer :
We know that the relation between the tangent and centre of the circle is that the perpendicular distance between the centre of the circle and the tangent of the circle is equal to radius of the circle.
The perpendicular distance between a line $ax+by+c=0$ and a point $\left( {{x}_{1}},{{y}_{1}} \right)$ is given by $\text{distance}=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$.
We are given a tangent equation where l and m are variables as $lx+my+1=0$ and a relation between l and m as $4{{l}^{2}}-5{{m}^{2}}+6l+1=0$. We should check every option with the known property of the tangent and the option which leads us to the relation $4{{l}^{2}}-5{{m}^{2}}+6l+1=0$ is the answer.
Let us consider option-A.
Centre = $\left( 2,0 \right)$
Radius = $3$
Using the distance formula where
$\left( {{x}_{1}},{{y}_{1}} \right)=\left( 2,0 \right)$
Distance = radius = 3
$ax+by+c=lx+my+1=0$

& 3=\dfrac{\left| l\times 2+m\times 0+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}} \\\ & \left| 2l+1 \right|=3\sqrt{{{l}^{2}}+{{m}^{2}}} \\\ \end{aligned}$$ Squaring on both sides, we get $${{\left( \left| 2l+1 \right| \right)}^{2}}={{\left( 3\sqrt{{{l}^{2}}+{{m}^{2}}} \right)}^{2}}$$ We know that ${{\left( \left| x \right| \right)}^{2}}={{x}^{2}}$ $\begin{aligned} & 4{{l}^{2}}+1+4l=9{{l}^{2}}+9{{m}^{2}} \\\ & 5{{l}^{2}}+9{{m}^{2}}-4l-1=0 \\\ \end{aligned}$ This end result is different from $4{{l}^{2}}-5{{m}^{2}}+6l+1=0$. So, option-A can’t be the answer. Let us consider option-B. Centre = $\left( -3,0 \right)$ Radius = $\sqrt{3}$ Using the distance formula where $\left( {{x}_{1}},{{y}_{1}} \right)=\left( -2,0 \right)$ Distance = radius = $\sqrt{3}$ $ax+by+c=lx+my+1=0$ $$\begin{aligned} & \sqrt{3}=\dfrac{\left| l\times \left( -3 \right)+m\times 0+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}} \\\ & \left| -3l+1 \right|=\sqrt{3}\sqrt{{{l}^{2}}+{{m}^{2}}} \\\ \end{aligned}$$ Squaring on both sides, we get $${{\left( \left| -3l+1 \right| \right)}^{2}}={{\left( \sqrt{3}\sqrt{{{l}^{2}}+{{m}^{2}}} \right)}^{2}}$$ $\begin{aligned} & 9{{l}^{2}}+1-6l=3{{l}^{2}}+3{{m}^{2}} \\\ & 6{{l}^{2}}-3{{m}^{2}}-6l+1=0 \\\ \end{aligned}$ This end result is different from $4{{l}^{2}}-5{{m}^{2}}+6l+1=0$. So, option-B can’t be the answer. Let us consider option-C. Centre = $\left( 3,0 \right)$ Radius = $\sqrt{5}$ Using the distance formula where $\left( {{x}_{1}},{{y}_{1}} \right)=\left( 3,0 \right)$ Distance = radius = $\sqrt{5}$ $ax+by+c=lx+my+1=0$ $$\begin{aligned} & \sqrt{5}=\dfrac{\left| l\times 3+m\times 0+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}} \\\ & \left| 3l+1 \right|=\sqrt{5}\sqrt{{{l}^{2}}+{{m}^{2}}} \\\ \end{aligned}$$ Squaring on both sides, we get $${{\left( \left| 3l+1 \right| \right)}^{2}}={{\left( \sqrt{5}\sqrt{{{l}^{2}}+{{m}^{2}}} \right)}^{2}}$$ $\begin{aligned} & 9{{l}^{2}}+1+6l=5{{l}^{2}}+5{{m}^{2}} \\\ & 4{{l}^{2}}-5{{m}^{2}}+6l+1=0 \\\ \end{aligned}$ This end result is same as the given condition $4{{l}^{2}}-5{{m}^{2}}+6l+1=0$. So, we can conclude that option-C is the answer. $\therefore $ $lx+my+1=0$ can be a tangent to the circle with centre $\left( 3,0 \right)$ and radius $\sqrt{5}$. The answer is option-C. **Note** :Another way to do this question is by directly applying the perpendicular relation by considering a centre $\left( {{x}_{1}},{{y}_{1}} \right)$ and radius as r units. This will lead to $\begin{aligned} & r=\dfrac{\left| l{{x}_{1}}+m{{y}_{1}}+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}} \\\ & \left| l{{x}_{1}}+m{{y}_{1}}+1 \right|=r\sqrt{{{l}^{2}}+{{m}^{2}}} \\\ \end{aligned}$ By squaring this equation and comparing it with $4{{l}^{2}}-5{{m}^{2}}+6l+1=0$, we get the answer. ${{\left( \left| l{{x}_{1}}+m{{y}_{1}}+1 \right| \right)}^{2}}={{\left( r\sqrt{{{l}^{2}}+{{m}^{2}}} \right)}^{2}}$ We know that ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ca \right)$ ${{l}^{2}}{{x}_{1}}^{2}+{{m}^{2}}{{y}_{1}}^{2}+1+2\left( lm{{x}_{1}}{{y}_{1}}+m{{y}_{1}}+l{{x}_{1}} \right)={{r}^{2}}\left( {{l}^{2}}+{{m}^{2}} \right)$ We can infer that there is no m term in the equation $4{{l}^{2}}-5{{m}^{2}}+6l+1=0$ which tells us that ${{y}_{1}}=0$ in the above equation. $\begin{aligned} & {{l}^{2}}{{x}_{1}}^{2}+1+2l{{x}_{1}}={{r}^{2}}\left( {{l}^{2}}+{{m}^{2}} \right) \\\ & {{l}^{2}}\left( {{x}_{1}}^{2}-{{r}^{2}} \right)-{{r}^{2}}{{m}^{2}}+2l{{x}_{1}}+1=0 \\\ \end{aligned}$ By comparing the coefficients, we get $\begin{aligned} & 2{{x}_{1}}=6 \\\ & {{x}_{1}}=3 \\\ \end{aligned}$ $\begin{aligned} & {{r}^{2}}=5 \\\ & r=\sqrt{5} \\\ \end{aligned}$ So, the centre is $\left( 3,0 \right)$ and radius is $\sqrt{5}$.