Question

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

x^2 + (y - k)^2 = 21.
What is the value of k in this equation?

Answer 1

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