SOLUTION: Show algebraically that the line 3x+2y=-4 passes through the center of the circle given by the equation x^2+y^2-4x+10y+19=0. HELP:)

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Question 139720: Show algebraically that the line 3x+2y=-4 passes through the center of the circle given by the equation x^2+y^2-4x+10y+19=0. HELP:)
Answer by Earlsdon(6103) About Me  (Show Source):
You can put this solution on YOUR website!
First, to find the center of the circle, you want to get your equation into the standard form for a circle with center at (h, k) and radius, r.
%28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2
To do this, you must "complete the square" in the x- and y-terms of your equation.
x%5E2%2By%5E2-4x%2B10y%2B19+=+0 Group the x- and y-terms together.
%28x%5E2-4x%29%2B%28y%5E2%2B10y%29%2B19+=+0 Now subtract 19 from both sides.
%28x%5E2-4x%29%2B%28y%5E2%2B10y%29+=+-19 Next, complete the squares in the x- and y-terms by adding the square of half the x-coefficient and the square of half the y-coefficient to both sides of the equation.
%28x%5E2-4x%2B4%29%2B%28y%5E2%2B10y%2B25%29+=+4%2B25-19 Factor the x- and y-terms on the left side and simplify the right side.
%28x-2%29%5E2%2B%28y%2B5%29%5E2+=+10 now compare this with the general form:
%28x-h%29%5E2%2B%28y-k%29%5E2+=+r%5E2...and you can see that the center of the circle, (h, k) is (2, -5)
Now you need to see if the point, (2, -5) satisfies the given equation:
3x%2B2y+=+-4 Substitute x = 2, and y = -5.
3%282%29%2B2%28-5%29+=+-4 Simplify.
6%2B%28-10%29+=+-4
-4+=+-4 So the answer is yes, the line does pass through the center of the circle.!