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)#
