If $4{{l}^{2}}-5{{m}^{2}}+6l+1=0,$ then show that line $lx+my+1=0$ touches a fixed circle. Find the equation of the circle.
Answer
329.7k+ views
Hint: Assume a circle and apply the condition to tangency on the given line to this assumed circle. Compare this equation with the equation given in the question.
Complete step by step solution:
Let us consider a circle having its equation ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{a}^{2}}$. The circle is in its standard form. So its centre is $\left( h,k \right)$ and radius $=a$.
Since line \[lx+my+1=0\] touches this circle, the perpendicular distance from centre of the circle ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{a}^{2}}$ to the line \[lx+my+1=0\] is equal to radius.
To find the perpendicular distance from a point $\left( {{x}_{1}},{{y}_{1}} \right)$ to a line \[ax+by+c=0\], the formula to find this perpendicular distance $'d'$ is given by
$d=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}........\left( i \right)$
Using formula $\left( i \right)$ to find perpendicular distance by substituting $a=l\text{,}b=m,c=1\text{, }{{x}_{1}}=h,{{y}_{1}}=k$ in equation $\left( i \right)$, we get 🡪
$d=\dfrac{\left| lh+mk+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}}$
As explained in the above paragraph, this distance should be equal to the radius of the circle.
So, $d=a$
Since we have found that $d=\dfrac{\left| lh+mk+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}}$, hence 🡪
$\dfrac{\left| lh+mk+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}}=a$
Squaring both sides, we get 🡪
\[\dfrac{{{\left( lh+mk+1 \right)}^{2}}}{{{l}^{2}}+{{m}^{2}}}={{a}^{2}}\]
$\Rightarrow {{l}^{2}}{{h}^{2}}+{{m}^{2}}{{k}^{2}}+1+2lh+2mk+2lhmk={{a}^{2}}\left( {{l}^{2}}+{{m}^{2}} \right)$
$\Rightarrow {{l}^{2}}{{h}^{2}}+{{m}^{2}}{{k}^{2}}+1+2lh+2mk+2lhmk={{a}^{2}}{{l}^{2}}+{{a}^{2}}{{m}^{2}}$
$\Rightarrow \left( {{l}^{2}}{{h}^{2}}-{{a}^{2}}{{l}^{2}} \right)+\left( {{m}^{2}}{{k}^{2}}-{{a}^{2}}{{m}^{2}} \right)+1+2lh+2mk+2lhmk=$
$\Rightarrow {{l}^{2}}\left( {{h}^{2}}-{{a}^{2}} \right)+{{m}^{2}}\left( {{k}^{2}}-{{a}^{2}} \right)+2hklm+2hl+2km+1=0.............\left( ii \right)$
It is given in question;
$4{{l}^{2}}-5{{m}^{2}}+6l+1=0......\left( iii \right)$
So we have to compare the equations $\left( ii \right)$ and $\left( iii \right)$. To compare two equations, we have to equate the ratio of coefficient of same and common variable in the two equations.
I)Equating ratio of coefficients of \[l\] in equations $\left( ii \right)$ and $\left( iii \right)$ to the ratio of constant terms in equations $\left( ii \right)$ and $\left( iii \right)$ 🡪
$\begin{align}
& \dfrac{2h}{6}=\dfrac{1}{1} \\
& \Rightarrow h=\dfrac{6}{2} \\
& \Rightarrow h=3 \\
\end{align}$
II) Equating ratio of coefficients of \[m\] in equations $\left( ii \right)$ and $\left( iii \right)$ to the ratio
of constant terms in equations $\left( ii \right)$ and $\left( iii \right)$ 🡪$\begin{align}
& \left( \dfrac{2k}{0} \right)=\dfrac{1}{1} \\
& \Rightarrow k=0 \\
\end{align}$
III)Equating ratio of coefficients of $'{{l}^{2}}'$ in equations $\left( ii \right)$ and $\left( iii \right)$ to the ratio of constant terms in equations $\left( ii \right)$ and $\left( iii \right)$ 🡪
$\begin{align}
& \dfrac{{{h}^{2}}-{{a}^{2}}}{4}=\dfrac{1}{1} \\
& \Rightarrow {{h}^{2}}-{{a}^{2}}=4 \\
& \Rightarrow {{\left( 3 \right)}^{2}}-{{a}^{2}}=4 \\
& \Rightarrow {{a}^{2}}=9-4 \\
& \Rightarrow {{a}^{2}}=5 \\
& \Rightarrow a=\sqrt{5} \\
\end{align}$
Since there exists a single value of \[h,\text{ }k,\text{ }a\], the circle is a fixed circle. Also, substituting
\[h=3,k=0,a=\sqrt{5}\] in the assumed circle’s equation i.e. ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{a}^{2}}$, we get the equation of circle 🡪
${{\left( x-3 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\sqrt{5}}^{2}}$
${{\left( x-3 \right)}^{2}}+{{y}^{2}}=5$
Note: There is a possibility of committing mistakes while equating the coefficient ratio. So, in order to avoid this mistake, first collect all the terms of which we have to find the coefficient and then take the coefficient out of that term.
Let us consider a circle having its equation ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{a}^{2}}$. The circle is in its standard form. So its centre is $\left( h,k \right)$ and radius $=a$.
Since line \[lx+my+1=0\] touches this circle, the perpendicular distance from centre of the circle ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{a}^{2}}$ to the line \[lx+my+1=0\] is equal to radius.
To find the perpendicular distance from a point $\left( {{x}_{1}},{{y}_{1}} \right)$ to a line \[ax+by+c=0\], the formula to find this perpendicular distance $'d'$ is given by
$d=\dfrac{\left| a{{x}_{1}}+b{{y}_{1}}+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}........\left( i \right)$
Using formula $\left( i \right)$ to find perpendicular distance by substituting $a=l\text{,}b=m,c=1\text{, }{{x}_{1}}=h,{{y}_{1}}=k$ in equation $\left( i \right)$, we get 🡪
$d=\dfrac{\left| lh+mk+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}}$
As explained in the above paragraph, this distance should be equal to the radius of the circle.
So, $d=a$
Since we have found that $d=\dfrac{\left| lh+mk+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}}$, hence 🡪
$\dfrac{\left| lh+mk+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}}=a$
Squaring both sides, we get 🡪
\[\dfrac{{{\left( lh+mk+1 \right)}^{2}}}{{{l}^{2}}+{{m}^{2}}}={{a}^{2}}\]
$\Rightarrow {{l}^{2}}{{h}^{2}}+{{m}^{2}}{{k}^{2}}+1+2lh+2mk+2lhmk={{a}^{2}}\left( {{l}^{2}}+{{m}^{2}} \right)$
$\Rightarrow {{l}^{2}}{{h}^{2}}+{{m}^{2}}{{k}^{2}}+1+2lh+2mk+2lhmk={{a}^{2}}{{l}^{2}}+{{a}^{2}}{{m}^{2}}$
$\Rightarrow \left( {{l}^{2}}{{h}^{2}}-{{a}^{2}}{{l}^{2}} \right)+\left( {{m}^{2}}{{k}^{2}}-{{a}^{2}}{{m}^{2}} \right)+1+2lh+2mk+2lhmk=$
$\Rightarrow {{l}^{2}}\left( {{h}^{2}}-{{a}^{2}} \right)+{{m}^{2}}\left( {{k}^{2}}-{{a}^{2}} \right)+2hklm+2hl+2km+1=0.............\left( ii \right)$
It is given in question;
$4{{l}^{2}}-5{{m}^{2}}+6l+1=0......\left( iii \right)$
So we have to compare the equations $\left( ii \right)$ and $\left( iii \right)$. To compare two equations, we have to equate the ratio of coefficient of same and common variable in the two equations.
I)Equating ratio of coefficients of \[l\] in equations $\left( ii \right)$ and $\left( iii \right)$ to the ratio of constant terms in equations $\left( ii \right)$ and $\left( iii \right)$ 🡪
$\begin{align}
& \dfrac{2h}{6}=\dfrac{1}{1} \\
& \Rightarrow h=\dfrac{6}{2} \\
& \Rightarrow h=3 \\
\end{align}$
II) Equating ratio of coefficients of \[m\] in equations $\left( ii \right)$ and $\left( iii \right)$ to the ratio
of constant terms in equations $\left( ii \right)$ and $\left( iii \right)$ 🡪$\begin{align}
& \left( \dfrac{2k}{0} \right)=\dfrac{1}{1} \\
& \Rightarrow k=0 \\
\end{align}$
III)Equating ratio of coefficients of $'{{l}^{2}}'$ in equations $\left( ii \right)$ and $\left( iii \right)$ to the ratio of constant terms in equations $\left( ii \right)$ and $\left( iii \right)$ 🡪
$\begin{align}
& \dfrac{{{h}^{2}}-{{a}^{2}}}{4}=\dfrac{1}{1} \\
& \Rightarrow {{h}^{2}}-{{a}^{2}}=4 \\
& \Rightarrow {{\left( 3 \right)}^{2}}-{{a}^{2}}=4 \\
& \Rightarrow {{a}^{2}}=9-4 \\
& \Rightarrow {{a}^{2}}=5 \\
& \Rightarrow a=\sqrt{5} \\
\end{align}$
Since there exists a single value of \[h,\text{ }k,\text{ }a\], the circle is a fixed circle. Also, substituting
\[h=3,k=0,a=\sqrt{5}\] in the assumed circle’s equation i.e. ${{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}={{a}^{2}}$, we get the equation of circle 🡪
${{\left( x-3 \right)}^{2}}+{{\left( y-0 \right)}^{2}}={{\sqrt{5}}^{2}}$
${{\left( x-3 \right)}^{2}}+{{y}^{2}}=5$
Note: There is a possibility of committing mistakes while equating the coefficient ratio. So, in order to avoid this mistake, first collect all the terms of which we have to find the coefficient and then take the coefficient out of that term.
Last updated date: 01st Jun 2023
•
Total views: 329.7k
•
Views today: 7.86k
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
