Question

For an equation of circle find its centre and radius when the circle has $lx+my+1=0$ as a tangent and also the relation $4{{l}^{2}}-5{{m}^{2}}+6l+1=0$ satisfies.

Answer 1

Hint:We try to form the equation of the circle in the form of general equation of the conic section. From the equation, we get the modulus form to find the radius of the circle. We then convert the equation in the distance formula to find the center of the circle. The center coordinates will be found from the numerator the distance formula.

Complete step by step answer:
We know that for a given relation of two unknown which has a linear combination as a tangent of a circle we can find out the equation of the circle from that circle.
Here the given relation is $4{{l}^{2}}-5{{m}^{2}}+6l+1=0$.
We change the equation to form the equation of circle.
$\begin{align}
  & 4{{l}^{2}}-5{{m}^{2}}+6l+1=0 \\
 & \Rightarrow 4{{l}^{2}}+5{{l}^{2}}+6l+1=5{{l}^{2}}+5{{m}^{2}} \\
 & \Rightarrow 9{{l}^{2}}+6l+1=5\left( {{l}^{2}}+{{m}^{2}} \right) \\
 & \Rightarrow {{\left( 3l+1 \right)}^{2}}=5\left( {{l}^{2}}+{{m}^{2}} \right) \\
\end{align}$
Now we try to get to the modulus form to get the radius of the circle.
So,
$\begin{align}
  & {{\left( 3l+1 \right)}^{2}}=5\left( {{l}^{2}}+{{m}^{2}} \right) \\
 & \Rightarrow \dfrac{{{\left( 3l+1 \right)}^{2}}}{\left( {{l}^{2}}+{{m}^{2}} \right)}=5 \\
 & \Rightarrow \left| \dfrac{\left( 3l+1 \right)}{\sqrt{\left( {{l}^{2}}+{{m}^{2}} \right)}} \right|=\sqrt{5} \\
\end{align}$
Here, $\sqrt{5}$ is the radius of the circle. This equation of circle is derived from the general equation of conic section $A{{x}^{2}}+Bxy+C{{y}^{2}}+Dx+Ey+F=0$. It becomes an equation of circle only when ${{B}^{2}}<4AC$.
The equation also formed with the equation of finding distance of the perimeter point from the centre. So, the equation $\left| \dfrac{\left( 3l+1 \right)}{\sqrt{\left( {{l}^{2}}+{{m}^{2}} \right)}} \right|=\sqrt{5}$ represents the centre at its numerator.
So, the centre is $3l+0\times m+1$ form of $al+bm+c$.
Equating we get that the centre is $\left( 3,0 \right)$.
The equation of circle has radius of $\sqrt{5}$ unit and centre being $\left( 3,0 \right)$.

Note:
We need to remember that the equation will be found not in its normal form. We also don’t need to convert it into the normal form which is unnecessary.
From the general form of conic section, we can find the centre and the radius of the circle in the distance formula. The distance formula for general is $\left| \dfrac{\left( al+bm+c \right)}{\sqrt{\left( {{l}^{2}}+{{m}^{2}} \right)}} \right|$ when the line is $xl+ym+c=0$.