Question 1119396
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<pre>
    x^2 + y^2 - 4x + 6y - 12 = 0      <<<---=== re-group to get


    (x^2 - 4x) + (y^2 + 6y) = 12      <<<---=== make identical transformations as shown below


    (x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9    <<<---=== complete the squares


    (x-2)^2 + (y+3)^2 = 25            <<<---===  Standard equation of the circle


Your circle has the center at (2,-3) and the radius  {{{sqrt(25)}}} = 5.


The distance from the center to the given point (4,-1) is

    d = {{{sqrt((4-2)^2 + ((-1)-(-3))^2)}}} = {{{sqrt(2^2 + 2^2)}}} = {{{sqrt(8)}}}


which is less than 5.


Hence, the point (4,-1) lies INSIDE the given circle.


Therefore, any line passing through (4,-1) cannot be tangent to the circle.
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Answered and solved.