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I need help with - Co-ordinate geometry - JEE Main

The equation of the circle, which is the mirror image of the circle, x^{2}+y^{2}-2x=0,in\; the\; line,\; y=3-x\; is:

  • Option 1)

    x^{2}+y^{2}-6x-4y+12=0

  • Option 2)

    x^{2}+y^{2}-6x-8y+24=0

  • Option 3)

    x^{2}+y^{2}-8x-6y+24=0

  • Option 4)

    x^{2}+y^{2}-4x-6y+12=0

 
Answers (1)
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As learnt in

Equation of a circle -

\left ( x-h \right )^{2}+\left ( y-k \right )^{2}= r^{2}

- wherein

Circle with centre \left ( h,k \right ) and radius r.

 

 

General form of a circle -

x^{2}+y^{2}+2gx+2fy+c= 0
 

- wherein

centre = \left ( -g,-f \right )

radius = \sqrt{g^{2}+f^{2}-c}

 Centre: (1,0) radius = 1

Image of A in the line

x+y=3.

Equation of AB is

x-y=1

Coordinate of C: x+y=3

                           x-y=1

                         ------------

                       x=2, y=1

\frac{h+1}{2}=2,\:\frac{k}{2}=1

h=3, k=2

Coordinates of B is (3,2) which is mirror image of centre (1,0) and radius is same =1

Hence equation is (x-3)2 + (y-2)2 = 1

x2+y2-6x-4y+12=0


Option 1)

x^{2}+y^{2}-6x-4y+12=0

This option is correct.

Option 2)

x^{2}+y^{2}-6x-8y+24=0

This option is incorrect.

Option 3)

x^{2}+y^{2}-8x-6y+24=0

This option is incorrect.

Option 4)

x^{2}+y^{2}-4x-6y+12=0

This option is incorrect.

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