Answer
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Hint: Equation of family of circle is${{S}_{1}}+\lambda {{S}_{2}}=0$. So, there can be infinite number of circles passing through the intersection points of ${{x}^{2}}+{{y}^{2}}=4$and ${{x}^{2}}+{{y}^{2}}-2x-4y+4=0$. By substituting values in the Family of circle equation we get the centre of the circle. Hence, we can calculate radius using $\sqrt{{{\left( x \right)}^{2}}+{{\left( y \right)}^{2}}-c}$ where x and y are coordinates and c is the constant in the equation.
Then find the perpendicular distance between radius and tangent to get the value of $\lambda $.
Then substitute this value in the equation of the family of circles.
Complete step by step answer:
Let us consider,
${{S}_{1}}={{x}^{2}}+{{y}^{2}}-4$ and ${{S}_{2}}={{x}^{2}}+{{y}^{2}}-2x-4y+4$
Now substitute this value in the family of circles equation.
$\begin{align}
& {{S}_{1}}+\lambda {{S}_{2}}=0 \\
& {{x}^{2}}+{{y}^{2}}-2x-4y+4+\lambda \left( {{x}^{2}}+{{y}^{2}}-4 \right)=0 \\
\end{align}$
$\left( \lambda +1 \right){{x}^{2}}+\left( \lambda +1 \right){{y}^{2}}-2x-4y-4\lambda +4=0......(1)$
Now divide the whole equation by $\left( \lambda +1 \right)$,
$\left( \lambda +1 \right){{x}^{2}}+\left( \lambda +1 \right){{y}^{2}}-2x-4y-4\lambda +4=0$
${{x}^{2}}+{{y}^{2}}-\dfrac{2x}{\left( \lambda +1 \right)}-\dfrac{4y}{\left( \lambda +1 \right)}+\dfrac{4-4\lambda }{\left( \lambda +1 \right)}=0$
Now compare this equation with the general equation of circle $a{{x}^{2}}+b{{y}^{2}}+2gx+2fy+c=0$
Coordinates of centre are given by $\left( g,f \right)$
Therefore, coordinates of the circle will be $\left( \dfrac{1}{\lambda +1},\dfrac{2}{\lambda +1} \right)$
Radius is obtained by $\sqrt{{{\left( g \right)}^{2}}+{{\left( f \right)}^{2}}-\left( c \right)}$
Now substitute the values and find radius.
$\sqrt{{{\left( \dfrac{1}{\lambda +1} \right)}^{2}}+{{\left( \dfrac{2}{\lambda +1} \right)}^{2}}-\left( \dfrac{-4\lambda +4}{\lambda +1} \right)}$
$\sqrt{\left( \dfrac{1+4-\left( -4\lambda +4 \right)\left( \lambda +1 \right)}{{{\left( \lambda +1 \right)}^{2}}} \right)}$
$\sqrt{\left( \dfrac{5-\left( 4+4\lambda -4\lambda -4{{\lambda }^{2}} \right)}{{{\left( \lambda +1 \right)}^{2}}} \right)}$
$\sqrt{\left( \dfrac{5-4+4{{\lambda }^{2}}}{{{\left( \lambda +1 \right)}^{2}}} \right)}$
$\sqrt{\left( \dfrac{1+4{{\lambda }^{2}}}{{{\left( \lambda +1 \right)}^{2}}} \right)}$
As, our circle is touching $x+2y=0$ i.e. this line is a tangent to the circle.
We know that the tangent is perpendicular to the radius.
And the perpendicular distance is given by
Radius = $\left| \dfrac{g+2f}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|$ where a and b are the coordinates of the perpendicular.
Now substitute the values,
$\sqrt{\left( \dfrac{1+4{{\lambda }^{2}}}{{{\left( \lambda +1 \right)}^{2}}} \right)}=\left| \dfrac{\left( \dfrac{1}{\lambda +1} \right)+\left( \dfrac{4}{\lambda +1} \right)}{\sqrt{{{\left( 1 \right)}^{2}}+{{\left( 2 \right)}^{2}}}} \right|$
Squaring both the sides to remove root as well as modulus.
$\begin{align}
& \sqrt{\left( \dfrac{1+4{{\lambda }^{2}}}{{{\left( \lambda +1 \right)}^{2}}} \right)}=\left| \dfrac{\left( \dfrac{5}{\lambda +1} \right)}{\sqrt{{{\left( 1 \right)}^{2}}+{{\left( 2 \right)}^{2}}}} \right| \\
& \left( \dfrac{1+4{{\lambda }^{2}}}{{{\left( \lambda +1 \right)}^{2}}} \right)=\dfrac{25}{{{\left( \lambda +1 \right)}^{2}}\times 5} \\
& 1+4{{\lambda }^{2}}=5 \\
& 4{{\lambda }^{2}}=4 \\
& \lambda =\pm 1 \\
\end{align}$
But $\lambda $cannot be zero.
Hence, the value of $\lambda $ will be 1.
Therefore, the co-ordinates will be $\left( \dfrac{1}{2},\dfrac{2}{2} \right)$
Substitute these values in this$\left( \lambda +1 \right){{x}^{2}}+\left( \lambda +1 \right){{y}^{2}}-2x-4y+4-4\lambda =0$ equation.
Hence, the equation of our required circle will be,
$\begin{align}
& \left( 1+1 \right){{x}^{2}}+\left( 1+1 \right){{y}^{2}}-2x-4y+4-4\left( 1 \right)=0 \\
& 2{{x}^{2}}+2{{y}^{2}}-2x-4y+4-4=0 \\
& {{x}^{2}}+{{y}^{2}}-x-2y=0 \\
\end{align}$
Hence, equation of circle is ${{x}^{2}}+{{y}^{2}}-x-2y=0$.
Note: Remember that the value of is to be taken positive. The formula of radius is $\sqrt{{{\left( g \right)}^{2}}+{{\left( f \right)}^{2}}-\left( c \right)}$ g, f being the centre of circle and c being the constant in the equation. In the formula of the perpendicular distance applying mod is necessary.
Then find the perpendicular distance between radius and tangent to get the value of $\lambda $.
Then substitute this value in the equation of the family of circles.
Complete step by step answer:
Let us consider,
${{S}_{1}}={{x}^{2}}+{{y}^{2}}-4$ and ${{S}_{2}}={{x}^{2}}+{{y}^{2}}-2x-4y+4$
Now substitute this value in the family of circles equation.
$\begin{align}
& {{S}_{1}}+\lambda {{S}_{2}}=0 \\
& {{x}^{2}}+{{y}^{2}}-2x-4y+4+\lambda \left( {{x}^{2}}+{{y}^{2}}-4 \right)=0 \\
\end{align}$
$\left( \lambda +1 \right){{x}^{2}}+\left( \lambda +1 \right){{y}^{2}}-2x-4y-4\lambda +4=0......(1)$
Now divide the whole equation by $\left( \lambda +1 \right)$,
$\left( \lambda +1 \right){{x}^{2}}+\left( \lambda +1 \right){{y}^{2}}-2x-4y-4\lambda +4=0$
${{x}^{2}}+{{y}^{2}}-\dfrac{2x}{\left( \lambda +1 \right)}-\dfrac{4y}{\left( \lambda +1 \right)}+\dfrac{4-4\lambda }{\left( \lambda +1 \right)}=0$
Now compare this equation with the general equation of circle $a{{x}^{2}}+b{{y}^{2}}+2gx+2fy+c=0$
Coordinates of centre are given by $\left( g,f \right)$
Therefore, coordinates of the circle will be $\left( \dfrac{1}{\lambda +1},\dfrac{2}{\lambda +1} \right)$
Radius is obtained by $\sqrt{{{\left( g \right)}^{2}}+{{\left( f \right)}^{2}}-\left( c \right)}$
Now substitute the values and find radius.
$\sqrt{{{\left( \dfrac{1}{\lambda +1} \right)}^{2}}+{{\left( \dfrac{2}{\lambda +1} \right)}^{2}}-\left( \dfrac{-4\lambda +4}{\lambda +1} \right)}$
$\sqrt{\left( \dfrac{1+4-\left( -4\lambda +4 \right)\left( \lambda +1 \right)}{{{\left( \lambda +1 \right)}^{2}}} \right)}$
$\sqrt{\left( \dfrac{5-\left( 4+4\lambda -4\lambda -4{{\lambda }^{2}} \right)}{{{\left( \lambda +1 \right)}^{2}}} \right)}$
$\sqrt{\left( \dfrac{5-4+4{{\lambda }^{2}}}{{{\left( \lambda +1 \right)}^{2}}} \right)}$
$\sqrt{\left( \dfrac{1+4{{\lambda }^{2}}}{{{\left( \lambda +1 \right)}^{2}}} \right)}$
As, our circle is touching $x+2y=0$ i.e. this line is a tangent to the circle.
We know that the tangent is perpendicular to the radius.
And the perpendicular distance is given by
Radius = $\left| \dfrac{g+2f}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right|$ where a and b are the coordinates of the perpendicular.
Now substitute the values,
$\sqrt{\left( \dfrac{1+4{{\lambda }^{2}}}{{{\left( \lambda +1 \right)}^{2}}} \right)}=\left| \dfrac{\left( \dfrac{1}{\lambda +1} \right)+\left( \dfrac{4}{\lambda +1} \right)}{\sqrt{{{\left( 1 \right)}^{2}}+{{\left( 2 \right)}^{2}}}} \right|$
Squaring both the sides to remove root as well as modulus.
$\begin{align}
& \sqrt{\left( \dfrac{1+4{{\lambda }^{2}}}{{{\left( \lambda +1 \right)}^{2}}} \right)}=\left| \dfrac{\left( \dfrac{5}{\lambda +1} \right)}{\sqrt{{{\left( 1 \right)}^{2}}+{{\left( 2 \right)}^{2}}}} \right| \\
& \left( \dfrac{1+4{{\lambda }^{2}}}{{{\left( \lambda +1 \right)}^{2}}} \right)=\dfrac{25}{{{\left( \lambda +1 \right)}^{2}}\times 5} \\
& 1+4{{\lambda }^{2}}=5 \\
& 4{{\lambda }^{2}}=4 \\
& \lambda =\pm 1 \\
\end{align}$
But $\lambda $cannot be zero.
Hence, the value of $\lambda $ will be 1.
Therefore, the co-ordinates will be $\left( \dfrac{1}{2},\dfrac{2}{2} \right)$
Substitute these values in this$\left( \lambda +1 \right){{x}^{2}}+\left( \lambda +1 \right){{y}^{2}}-2x-4y+4-4\lambda =0$ equation.
Hence, the equation of our required circle will be,
$\begin{align}
& \left( 1+1 \right){{x}^{2}}+\left( 1+1 \right){{y}^{2}}-2x-4y+4-4\left( 1 \right)=0 \\
& 2{{x}^{2}}+2{{y}^{2}}-2x-4y+4-4=0 \\
& {{x}^{2}}+{{y}^{2}}-x-2y=0 \\
\end{align}$
Hence, equation of circle is ${{x}^{2}}+{{y}^{2}}-x-2y=0$.
Note: Remember that the value of is to be taken positive. The formula of radius is $\sqrt{{{\left( g \right)}^{2}}+{{\left( f \right)}^{2}}-\left( c \right)}$ g, f being the centre of circle and c being the constant in the equation. In the formula of the perpendicular distance applying mod is necessary.
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