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Let \mathrm{S_1 \equiv x^2+y^2-4 x-8 y+4=0} and \mathrm{S_2} be its image in the line \mathrm{y=x}. The equation of the circle touching \mathrm{y=x} at \mathrm{(1,1)}  and orthogonal to \mathrm{S_2} is
 

Option: 1

\mathrm{ x^2+y^2+x-5 y+2=0}


Option: 2

\mathrm{ x^2+y^2+5 x-y+2=0}


Option: 3

\mathrm{ x^2+y^2+x+2 y+1=0}


Option: 4

\mathrm{ x^2+y^2-2 x+y+1=0}


Answers (1)

best_answer

Centre of circle \mathrm{S}_1=(2,4)
Centre of circle \mathrm{S}_2=(4,2)
Radius of circle \mathrm{S}_1= radius of circle \mathrm{S}_2=4

\therefore equation of circle \mathrm{S}_2

(\mathrm{x}-4)^2+(\mathrm{y}-2)^2=16 \Rightarrow \mathrm{x}^2+\mathrm{y}^2-8 \mathrm{x}-4 \mathrm{y}+4=0\quad \quad \quad \quad \dots(i)
Equation of circle touching \mathrm{y}=\mathrm{x} at (1,1) can be taken as

(\mathrm{x}-1)^2+(\mathrm{y}-1)^2+\lambda(\mathrm{x}-\mathrm{y})=0
or, \mathrm{x}^2+\mathrm{y}^2+\mathrm{x}(\lambda-2)+\mathrm{y}(-\lambda-2)+2=0\quad \quad \quad \quad \dots(ii)

As this is orthogonal to \mathrm{S}_2

\begin{aligned} & \Rightarrow 2\left(\frac{\lambda-2}{2}\right) \cdot(-4)+2\left(\frac{-\lambda-2}{2}\right) \cdot(-2)=4+2 \Rightarrow-4 \lambda+8+2 \lambda+4=6 \\ & \Rightarrow \lambda=3 \\ & \therefore \text { Required equation of circle is } \\ & \mathrm{x}^2+\mathrm{y}^2+\mathrm{x}-5 \mathrm{y}+2=0 . \end{aligned}

\therefore Required equation of circle is
\mathrm{ x^2+y^2+x-5 y+2=0}
 

Posted by

sudhir.kumar

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