Question 139720
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.
{{{(x-h)^2+(y-k)^2 = r^2}}}
To do this, you must "complete the square" in the x- and y-terms of your equation.
{{{x^2+y^2-4x+10y+19 = 0}}} Group the x- and y-terms together.
{{{(x^2-4x)+(y^2+10y)+19 = 0}}} Now subtract 19 from both sides.
{{{(x^2-4x)+(y^2+10y) = -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.
{{{(x^2-4x+4)+(y^2+10y+25) = 4+25-19}}} Factor the x- and y-terms on the left side and simplify the right side.
{{{(x-2)^2+(y+5)^2 = 10}}} now compare this with the general form:
{{{(x-h)^2+(y-k)^2 = r^2}}}...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+2y = -4}}} Substitute x = 2, and y = -5.
{{{3(2)+2(-5) = -4}}} Simplify.
{{{6+(-10) = -4}}}
{{{-4 = -4}}} So the answer is yes, the line does pass through the center of the circle.!