Question

find the center and radlus of the circle x2+y2-8x-4y-5=0

Answer 1

Centre :(4,2)
Radius : (5 units)
Since, we don't have dimension given in question.
So, radius is in appropriate units.

Otherwise, it is weird setback.

Answer 2

$The \: \: answer \: \: is \: \: given \: \: below \\ \\ RULE \\ \\ The \: \: equation \: \: of \: \: a \: \: circle \: \: is\\ \\ {(x -a )}^{2} + {(y - b)}^{2} = {r}^{2},\\ \\ where \: \: (a,b) \: \: is \: \: the \: \: centre \: \: \\ of \: \: the \: \: circle \: \: and \: \: r \: \: is \: \: the \\ radius \: \: of \: \: the \: \: circle. \\ \\ SOLUTION \\ \\ The \: \: given \: \: equation \: \: of \: \: the \: \: \\ circle \: \: is \\ \\ {x}^{2} + {y}^{2} - 8x - 4y - 5 = 0 \\ \\ Or, \: \: ( {x}^{2} - 8x + 16) + ( {y}^{2} - 4y + 4) = 5 + 16 + 4 \\ \\ Or, \: \: {(x - 4)}^{2} + {(y - 2)}^{2} = {5}^{2} \\ \\ Hence, \: \: the \: \: centre \: \: of \: \: the \: \: given \\ circle \: \: is \: \: at \: \: (4,2) \: \: and \: \: the\\ radius \: \: of \: \: the \: \: circle \: \: is \\ 5 \: \: units. \\ \\ Thank \: \: you \: \: for \: \: the \: \: question.$