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The image of the circle  \mathrm{x^2+y^2-8 x-6 y-24=0}  in the line  \mathrm{2 x+3 y-4=0}  is

Option: 1

\mathrm{x^2+y^2+6 y-40=0 }


Option: 2

\mathrm{x^2+y^2-6 y-40=0 }


Option: 3

\mathrm{x^2+y^2+6 y+40=0 }


Option: 4

\mathrm{x^2+y^2-6 y+40=0}


Answers (1)

best_answer

Given circle is  \mathrm{x^2+y^2-8 x-6 y-24=0} 


\mathrm{ \therefore \quad C(4,3), r=7 }
Given line is   \mathrm{ 2 x+3 y-4=0 }
Radius of required circle = Radius of original (given) circle =7
Let (h, k) be centre of required circle. Then (h, k)= image of (4,3) in the mirror line

\mathrm{ \therefore \quad \frac{h-4}{2}=\frac{k-3}{3}=\frac{-2|2(4)+3(3)-4|}{2^2+3^2} \\ }
\mathrm{ \Rightarrow(h, k)=(0,-3) \\ }
\mathrm{ \therefore \quad \text { Required circle is }(x-0)^2+(y+3)^2=49 \\ }
\mathrm{ \Rightarrow x^2+y^2+6 y-40=0 }\mathrm{ \Rightarrow x^2+y^2+6 y-40=0 }
 

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Shailly goel

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