SOLUTION: Explain why any line passing through (4,-1) and cannot be tangent to the circle x^2 + y^2 - 4x + 6y - 12 = 0

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Question 1119396: Explain why any line passing through (4,-1) and cannot be tangent to the circle x^2 + y^2 - 4x + 6y - 12 = 0
Found 2 solutions by josmiceli, ikleyn:
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
The only way the line through (4,-1) could
not be tanghent to the circle is if this
point is inside the circle.
----------------------------
Get the equation into the form:

where the center is at ( h, k )

Complete the squares for both x and y
and add to the right side to balence equation


The center is at ( 2, -3 ) and the radius is
-------------------------------
I just have to find out how far ( 4, -1 ) is from the center





This distance is shorter than the radius, so
( 4, -1 ) is inside the circle and a line
passing through it cannot be tangent to the circle
------------------------------------------------
try to get a 2nd opinion on this. I definitely
could have made a mistake, but I think my method is OK


Answer by ikleyn(52879)   (Show Source): You can put this solution on YOUR website!
.
    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   = 5.


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

    d =  =  = 


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.


Answered and solved.


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