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Let x^(2)+y^(2)+2gx+2fy+c=0 be an equati...

Let `x^(2)+y^(2)+2gx+2fy+c=0` be an equation of circle. Match the following lists :

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The correct Answer is:
`a rarr p; r,s; b rarr r,s; c rarr q,s; d rarr p,s`
a. The point `(-g,-f)` lies in the first quadrant. Then `glt0` and `flt0` . Also , the x- and y-axes must not cut the circle.
Solving the circle and the x-axis, we have `x^(2)+2gx+c=-`, which must chave imaginary roots. So, `g^(2) - c lt0`. Then c must be positive . Also, `f^(2) - c lt0`
b. If the circle lies above the x-axis, then `x^(2)+2gx+c=0` must have imaginary roots. Therefore, `g^(2) -c lt 0 ` and `cgt0`.
c. The point `(-g,-f)` lies in the third or fourth quadrant. Then `ggt0`. Also, the y -axis must not cut the circle.
Solving the circle and the y-axis, we have `y^(2)+2gy +c=0`, which must have imaginary roots. Then `f^(2)-c lt0`. So, c must be positive.
d. `x^(2)+2gx=c=0` must have equal roots. Then `g^(2) =c`. Hence, `cgt0`.
Also, `-g gt 0` or `g lt0`.
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