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Question

The equation of the image of the circle $${ x }^{ 2 }+{ y }^{ 2 }+16x-24y+183=0$$ by the line mirror $$4x+7y+13=0$$


A
x2+y2+32x+4y+235=0
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B
x2+y232x+4y+235=0
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C
x2+y2+32x+4y235=0
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D
None of these
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Solution

The correct option is B $${ x }^{ 2 }+{ y }^{ 2 }+32x+4y+235=0$$
The given circle and line are 
$${ x }^{ 2 }+{ y }^{ 2 }+16x-24y+183=0$$     ...(1)
and, $$ 4x+7y+13=0$$      ...(2)
Centre and radius of circle (1) are $${ C }_{ 1 }\left( -8,-12 \right) $$ and $$5$$ respectively.
Let the centre of the imaged circle be $${ C }_{ 2 }\left( { x }_{ 1 },{ y }_{ 1 } \right) .$$
Then, slope of $${ C }_{ 1 }{ C }_{ 2 }\times $$ slope of given line 
$$\displaystyle \Rightarrow \frac { { y }_{ 1 }-12 }{ { x }_{ 1 }+8 } \times -\frac { 4 }{ 7 } =-1\Rightarrow 7{ x }_{ 1 }-4{ y }_{ 1 }+104=0$$   ...(3)
and mid point of $${ C }_{ 1 }{ C }_{ 2 }$$ i.e., $$\displaystyle \left( \frac { { x }_{ 1 }-8 }{ 2 } ,\frac { { y }_{ 1 }+12 }{ 2 }  \right) $$ lie on (2)
i.e., $$\displaystyle 4\left( \frac { { x }_{ 1 }-8 }{ 2 }  \right) +7\left( \frac { { y }_{ 1 }+12 }{ 2 }  \right) +13=0$$
or, $$4{ x }_{ 1 }+7{ y }_{ 1 }+78=0$$    (4)
Solving (3) and (4), we get $$\left( { x }_{ 1 },{ y }_{ 1 } \right) \equiv \left( -16,-2 \right) $$
$$\therefore$$ Equation of the imaged circle is 
$${ \left( x+16 \right)  }^{ 2 }+{ \left( y+2 \right)  }^{ 2 }={ 5 }^{ 2 }$$
$$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }+32x+4y+235=0.$$

390469_190698_ans.PNG

Physics

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