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Find the equation of circle passing through the intersection of ${{x}^{2}}+{{y}^{2}}=4$ and ${{x}^{2}}+{{y}^{2}}-2x-4y+4=0$ and touching the line $x+2y=0$

Last updated date: 16th Jun 2024
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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.
  & {{S}_{1}}+\lambda {{S}_{2}}=0 \\
 & {{x}^{2}}+{{y}^{2}}-2x-4y+4+\lambda \left( {{x}^{2}}+{{y}^{2}}-4 \right)=0 \\
$\left( \lambda +1 \right){{x}^{2}}+\left( \lambda +1 \right){{y}^{2}}-2x-4y-4\lambda +4=0......(1)$
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.
  & \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 \\
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,
  & \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 \\

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.