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.
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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
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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
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try to get a 2nd opinion on this. I definitely
could have made a mistake, but I think my method is OK
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.
