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The image of the circle \mathrm{x^2+y^2-14 x-8 y+1=0} in the line 2x+3 y=0 is

Option: 1

\mathrm{x^2+y^2+2 x+16 y+1=0}


Option: 2

\mathrm{x^2+y^2+2 x-16 y+1=0}


Option: 3

\mathrm{x^2+y^2+2 x+16 y+40=0}


Option: 4

\mathrm{x^2+y^2+2 x-16 y+40=0}


Answers (1)

best_answer

\mathrm{\text { : Given circle is } x^2+y^2-14 x-8 y+1=0}

∴ C(7, 4), r = 8
Given line is 2x + 3y = 0
Radius of required circle = Radius of original (given) circle = 8
Let (h, k) be centre of required circle. Then (h, k) = image of (7, 4) in the mirror line

\mathrm{\begin{aligned} & \therefore \quad \frac{h-7}{2}=\frac{k-4}{3}=\frac{-2|2(7)+3(4)|}{2^2+3^2} \\ & \Rightarrow(h, k)=(-1,-8) \end{aligned}}

\mathrm{\therefore \quad \text { Required circle is }(x+1)^2+(y+8)^2=64}

\mathrm{\Rightarrow x^2+y^2+2 x+16 y+1=0}

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