Ok, the main concept is to see that having a line tangent to a circle at a point P means that the radius passing by the point P is perpendicular to that line at that point.
Numerically, the line x-2y=-4 (same as y= x/2 + 2) is tangent to the circle at (0;2). Let's draw the radius going from the center of the circle to (0;2). We know that the line L1 containing this radius is perpendicular to y= x/2 + 2. So you can use the property that when two lines are perpendicular, the slope of the first line is equal to -1/(the slope of the second line). Cool huh ?
So the line y= x/2 + 2 has slope 1/2 so the slope of L1 is -1/(1/2) = -2. So the equation of L1 should be y1=-2x+b. And we know that is passes by (0;2) because it intesects the first line at that point so we put it in the equation :
y1=-2x+b so 2=-2 (0) + b so b=2. Great ! Now, y1=-2x+2.
Cool, we know that the center is somewhere on the line y1 but where ?
So it is here that we need the seconde tangent y=2x-7.
We use the same trick to find the second line L2 which contains the radius passing by (3;-1). Since it is perpendicular to y=2x-7, the slope of L2 is -1/(2) = -1/2. So again the equation of L2 is y2=-1/2 x + B. And we know that is passes by (3;-1) so
y2=-1/2 x + B so -1=-1/2 (3) + B so B=1/2. Ok y2=-1/2x+1/2.
So we know that the center of the circle (x,y) is the interection of L1 and L2 so at that point the two lines have the same x and the same y so we put
so -2x+2 = -1/2 x+1/2.
We multiply both sides by 2 : -4x+4 = -x+1.
We put the x together and the other numbers together : -3x=-3 so x = 1.
Now we find the y of the center by replacing the x=1 into one of the two equation y1 or y2 since they have the same y at that point it should be equal :
y1 = -2(1)+2 = 0 and y2=-1/2 (1)+1/2=0 Great ! Done
So the center is (1;0)