From this reference on Conic Sections we obtain the General Cartesian Form:

#Ax^2 + Bxy +Cy^2 + Dx + Ey + F = 0" [1]"#

Please observe in the given equation:

#x^2 + y^2 + 10x + 8y = 40" [2]"#

that #A = C and B = 0#. The reference tells us that this identifies a circle.

The Cartesian form for the equation of a circle is:

#(x-h)^2+(y-k)^2=r^2" [3]"#

where #(x,y)# is any point on the circle, #(h,k)# is the center point, and r is the radius.

I shall expand the squares of equation [3]:

#x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2" [4]"#

By matching the terms in equation [4] with equation [2], we observe that equation [2] does not have terms corresponding to #h^2# and #k^2#, therefore, we shall add #h^2 + k^2# to both sides of the equation and arrange the terms on the left side in the same order as equation [4]:

#x^2 + 10x+ h^2 + y^2 + 8y+ k^2 = 40+h^2+k^2" [5]"#

Matching terms in equation [4] with equation [5], we observe that the following must be true:

#-2hx = +10x# and #-2ky=+8y#

Dividing the equation by #-2x and -2y#, respectively, will allow us to discover the values of h and k:

#h = -5# and #k=-4#

It follows that:

#h^2 = 25# and #k^2 = 16#

Substitute the for #h^2 and k^2# values into equation [5]:

#x^2 + 10x+ 25 + y^2 + 8y+ 16 = 40+25+16" [6]"#

We know that the left side of equation [6] will collapse into squares in the same form as equation [3] with #h = -5 and k = -4#

#(x--5)^2 + (y--4)^2 = 40+25+16" [7]"#

Please notice that you can write the #--# as plus #+# but I discourage people from doing this, because it often causes errors, when identifying the center.

Simply the right side of equation [7] and write it as a square:

#(x--5)^2 + (y--4)^2 = 9^2" [8]"#

The domain is:

#-14 <=x<=4#

The range is:

#-13<=y<=5#

The circle will be symmetric about a line with any slope, m, of the form:

#y = m(x+5)-4#

The center is the point #(-5,-4)#