x²-y²-2x+4y-3=0
We complete the square. (This is a degenerate hyperbola
which is degenerated into its two asymptotes. Its center,
foci and vertices all coincide at the intersection of the
two asymptotes)
x²-2x-y²+4y=3
(x²-2x)-(y²-4y)=3
(x²-2x+1)-(y²-4y+4)=3+1-4
(x-1)²-(y-2)² = 0
Factor as the difference of two squares:
[(x-1)-(y-2)][(x-1)+(y-2)] = 0
[x-1-y+2][x-1+y-2] = 0
(x-y+1)(x+y-3) = 0
x-y+1 = 0; x+y-3 = 0
-y = -x-1; y = -x+3
y = x+1;
Those are the equations of the two lines.
We graph them and plot the point which the circle goes through
Since two diameters of the same circle must intersect at the center
of the circle. The center of the circle must be the intersection of
those two lines. So we sketch the circle below:
We solve them as a system to find their point
of intersection:
y = -x+3
y = x+1
-x+3 = x+1
-2x = -2
x = 1
Substitute
y = 1+1
y=2
So they intersect at (1,2), so that's the center, and the radius is 1
because the distance from the center to the point on the circle (1,1)
is 1 unit. So the equation is
(x-h)²+(y-k)² = r²
(x-1)²+(y-2)² = 1
If you like, you can multiply that out and get:
x²-2x+1+y²-2y+4 = 1
or
x²+y²-2x-2y+3 = 0
Edwin
