Question

If the equation of circle having the pair of lines ${{x}^{2}}+2xy+3x+6y=0$ as it’s normal and having the size just sufficient to contain the circle $x\left( x-4 \right)+y\left( y-3 \right)=0$ is given by ${{x}^{2}}+{{y}^{2}}+6x-3y-k=0$, then find the value of $k$.

Answer 1

For solving this type of question, we should know about the normal line concept in a good way. It consists of the intersection points and by this we will calculate our centre for the circle and then by this we can calculate the radius for the required circles.

Complete step by step answer:
According to our question, a circle is $x\left( x-4 \right)+y\left( y-3 \right)=0$. So, for calculating the radius of the circle $x\left( x-4 \right)+y\left( y-3 \right)=0$. BY solving this we will get,
$\begin{aligned} & \left( x-0 \right)\left( x-4 \right)+\left( y-0 \right)\left( y-3 \right)=0 \\\ & \because \left( {{x}_{1}},{{y}_{1}} \right)\left( {{x}_{2}},{{y}_{2}} \right)=\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)+\left( y-{{y}_{1}} \right)\left( y-{{y}_{2}} \right) \\\ \end{aligned}$
So, by this the centre of the circle is,
${{C}_{1}}\to \left( 0,0 \right)\left( 4,3 \right)$
Centre : $\left( 2,\dfrac{3}{2} \right)$
The radius of the circle is,
$\begin{aligned} & d=\sqrt{{{\left( 4-0 \right)}^{2}}+{{\left( 3-0 \right)}^{2}}} \\\ & \Rightarrow r=\dfrac{\sqrt{16+9}}{2}=\dfrac{\sqrt{25}}{2}=\dfrac{5}{2} \\\ \end{aligned}$
For the circle 2, we have,
$\begin{aligned} & {{x}^{2}}+2xy+3x+6y=0 \\\ & \Rightarrow x\left( x+2y \right)+3\left( x+2y \right)=0 \\\ & \Rightarrow \left( x+3 \right)\left( x+2y \right)=0 \\\ \end{aligned}$
So, the centre of the circle is,
$\begin{aligned} & x+3=0,x+2y=0 \\\ & \Rightarrow x=-3,2y=-x \\\ & \Rightarrow x=-3,y=\dfrac{3}{2} \\\ \end{aligned}$
So, the centre of the circle is,
${{C}_{2}}=\left( -3,\dfrac{3}{2} \right)$
So, the distance between both the circle centres is,
$\begin{aligned} & {{C}_{1}}{{C}_{2}}=R-r \\\ & \Rightarrow \sqrt{{{\left( 2-\left( -3 \right) \right)}^{2}}+{{\left( \dfrac{3}{2}-\dfrac{3}{2} \right)}^{2}}}=R-\dfrac{5}{2} \\\ \end{aligned}$
So, the value of $R$ is,
$\begin{aligned} & \sqrt{{{\left( 5 \right)}^{2}}}=R-\dfrac{5}{2} \\\ & \Rightarrow R=5+\dfrac{5}{2} \\\ & \Rightarrow R=\dfrac{15}{2} \\\ \end{aligned}$
So, the equation of the circle is,
$\begin{aligned} & {{\left( x+3 \right)}^{2}}+{{\left( y-\dfrac{3}{2} \right)}^{2}}={{\left( \dfrac{15}{2} \right)}^{2}} \\\ & \Rightarrow {{x}^{2}}+6x+9+{{y}^{2}}-3y+\dfrac{9}{4}=\dfrac{225}{4} \\\ \end{aligned}$
By solving this and multiplying by 4 on both sides we get,
$\begin{aligned} & 4{{x}^{2}}+24x+36+4{{y}^{2}}-12y+9=225 \\\ & \Rightarrow 4{{x}^{2}}+4{{y}^{2}}+24x-12y+180=0 \\\ \end{aligned}$
By dividing the whole equation by 4, we get,
${{x}^{2}}+{{y}^{2}}+6x-3y-45=0$
So, the value of $k$ by comparing both the equations is 45.

Note: When you solve this question then always first find the centre of the circle and then find the radius. And if one normal line is given, then apply it for both and if two circles are touching each other then they will definitely give some common values which will be used for solving the next circle or required circle.