Question

If the circle x² + y ² - 4 x-6y+k=0 does not
touch or intersect the coordinate axes
and the point (2,2) lies inside the circle-
If K belongs to interval (a,b) then (b-a) is​

Answer 1

Answer:

Answer is 3

Step-by-step explanation:

$given \: that \: the \: circle \: \\ {x}^{2} + {y}^{2} - 4x - 6y + k = 0 \\ does \: not \: touch \: any \: axis. \\ \\ centre \: of \: the \: circle \: (2 \: 3) \\ radius \: = \sqrt{4 + 9 - k} = \sqrt{13 - k} \\ \\ since \: circle \: does \: not \: touch \: axis \: hence \: radius \: \\ will \: be \: \\ \sqrt{13 - k} < 2 \: and \: \sqrt{13 - k} < 3 \\ k > 4 \: and \: \: k > 9.......(1) \\ \\ point \: (2 \:, 2) \: lie \: insde \: the \: circle \: hence \: \\ 4 + 4 - 8 - 12 + k < 0 \\ k < 12.......(2) \\ \\ from \: equation \: (1) \: and \: (2) \\ k \: lies \: in \: the \: interval \: (9 \:, 12) = (a \:, b) \\ \\ b - a = 12 - 9 = 3 \\ \\ hope \: it \: will \: help \: you....$