Question 141478
 Determine the equation of the line passing through the centres of the circles 
{{{(x-2)^2+(y+5)^2=16}}} and {{{3x^2+3y^2+24x-6y-24=0}}}

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We will first find the center of each circle and 
then find the equation of the line through them.

The first one is in standard form {{{(x-h)^2+(y-k)^2=r^2}}}
and the second one is in general form:{{{Ax^2+Ay^2+Dx+Ey+F=0}}}

The first equation is already in standard form

{{{(x-2)^2+(y+5)^2=16}}}

we compare it to:

{{{(x-h)^2+(y-k)^2=r^2}}}

So h=2, k=-5, r=4, and the center is (h,k) = (2,-5)

The second equation is not in standard form, so we
must get it in standard form also:

{{{3x^2+3y^2+24x-6y-24=0}}}

Divide every term by 3:

{{{x^2+y^2+8x-2y-8=0}}}

Get the x terms together, and the
y-terms together and put blanks
to add something to both sides

{{{x^2+8x+__+y^2-2y+___=8+__+___}}}

Complete the square on the x terms:

Take the coefficient of x, which is 8
Multiply it by {{{1/2}}}, getting 4
Square what you get, {{{4^2 = 16}}}
Add that in the first blanks on both sides

{{{x^2+8x+16+y^2-2y+___=8+16+___}}}

Complete the square on the y terms:

Take the coefficient of y, which is -2
Multiply it by {{{1/2}}}, getting -1
Square what you get, {{{(-1)^2 = 1}}}
Add that in the remaining blanks of both sides

{{{x^2+8x+16+y^2-2y+1=8+16+1}}}

Combine the terms on the right:

{{{x^2+8x+16+y^2-2y+1=25}}}

Factor the first three terms on the left as
a perfect square:

{{{(x+4)^2+(y-1)^2=25}}}

That is in standard form, so we compare it to:

{{{(x-h)^2+(y-k)^2=r^2}}}

So h=-4, k=1, r=5, and the center is (h,k) = (-4,1)

So now we just need to find the line passing through
the points (2,-5) and (-4,1)

Find the slope using

{{{m = (y2-y1)/(x2-x1)}}}

{{{m = ((1)-(-5))/((-4)-(2)) = (1+5)/(-4-2) = 6/(-6) = -1}}}

Now we use the point-slope form:

{{{y-y1=m(x-x1)}}}

{{{y-(-5))=-1(x-(2))}}}

{{{y+5=-(x-2)}}}

{{{y+5=-x+2}}}

{{{y=-x-3}}}

Plotting the circles and the line:

{{{drawing(400,400,-10,10,-10,10,
graph(400,400,-10,10,-10,10), circle(-4,1,5),line(15,-18,-15,12),circle(2,-5,4) )}}}

Edwin</pre>