Question

The circle given by : x^2 + y^2 - 6y - 12 = 0 can be written as:

Answer 1

Step-by-step explanation:

Given equation of circle is,

${x}^{2} + {y}^{2} - 6y - 12 = 0 \\ by \: compairing \: it \: with \: \\ {x}^{2} + {y}^{2} + 2gx + 2fy = 0 \: we \: get \\2 g = 0 \: or \: g = 0 \\ 2f= - 6 \\ or \: f= - 3 \\ c = - 12$

again the another form of circle is,

${x}^{2} + {(y - k)}^{2} = 21 \\ or \: {x}^{2} + {y}^{2} - 2ky + {k}^{2} - 21 = 0 \\ by \: compairing \:it \: with \: \\ {x}^{2} + {y}^{2} + 2gx + 2fy = 0 \: we \: get \\ g = 0 \\f = - k \\ c = {k}^{2} - 21$

now equating the values of f in both equations,

-k=-3

i.e. k=3

therefore k=3