Let the circle pass through point a (2, - 3), the center of the circle be on the straight line 2x + y = 0, and tangent to the straight line x-y-1 = 0, then the standard equation of the circle can be solved

Let the circle pass through point a (2, - 3), the center of the circle be on the straight line 2x + y = 0, and tangent to the straight line x-y-1 = 0, then the standard equation of the circle can be solved

Solution: let the coordinates of the center o be (a, - 2A), make a vertical line x-y-1 = 0 through O, and intersect with point B, then ob = OA,
The slope of the straight line x-y-1 = 0 is k = 1, and the slope of the vertical line k '= - 1 / k = - 1,
Let the equation of the vertical line be y = - x + B, and substitute it into the coordinates of O to get b = - A, that is, y = - x-a,
The simultaneous solution of the equation x-y-1 = 0 is: x = (- A-1) / 2, y = (1-A) / 2, that is, point B is ((- A-1) / 2, (1-A) / 2),
OB=v{[a-(-a-1)/2]^2+[-2a-(1-a)/2]^2}=v(3a+1)^2/2,
OA=v[(a-2)^2+(-2a+3)^2]=v(5a^2-16a+13),
OB = OA, a = 19 + - 4v21,
Radius r = ob = (3a + 1) / V2,
The standard equation of circle is: (x-a) ^ 2 + (y + 2a) ^ 2 = R ^ 2 = (3a + 1) ^ 2 / 2