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Please,please help me - Co-ordinate geometry - JEE Main-4

The point diametrically opposite to the point P(1,0)   on the  circle x^{2}+y^{2}+2x+4y-3=0  is

  • Option 1)

    (3, 4)

  • Option 2)

    (3, – 4)

  • Option 3)

    (– 3, 4)

  • Option 4)

    (– 3, – 4)

 
Answers (1)
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As we learnt in

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}

 

 

Mid-point formula -

x= \frac{x_{1}+x_{2}}{2}

y= \frac{y_{1}+y_{2}}{2}

 

- wherein

If the point P(x,y) is the mid point of line joining A(x1,y1) and B(x2,y2) .

 x^{2}+y^{2}+2x+4y-3=0

Centre is (-1, -2) 

\frac{x+1}{2}=-1\:;\:\:\frac{y+0}{2}=-2

x=-3\:;\:\:\y=-4


Option 1)

(3, 4)

This option is incorrect.

Option 2)

(3, – 4)

This option is incorrect.

Option 3)

(– 3, 4)

This option is incorrect.

Option 4)

(– 3, – 4)

This option is correct.

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