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

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 $${ \left( x+16 \right) }^{ 2 }+{ \left( y+2 \right) }^{ 2 }={ 5 }^{ 2 }$$$$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }+32x+4y+235=0.$$Physics

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