Given the circle m, the center of the circle is on the line 2x + y = 0, and tangent to the line x + Y-1 = 0 at the point a (2, - 1), find the standard equation of the circle m

Given the circle m, the center of the circle is on the line 2x + y = 0, and tangent to the line x + Y-1 = 0 at the point a (2, - 1), find the standard equation of the circle m

The circle is tangent to the line x + Y-1 = 0, that is, it is tangent to y = - x + 1, and the slope of the line y = - x + 1 is - 1. That is, the center of the circle should be on the vertical line of the line passing through point a. then the slope of the vertical line should be the opposite number of the slope of the line y = - x + 1, that is, 1
The tangent point is a (2, - 1), which is also a point on the vertical line. When the slope is known, the equation of the vertical line is obtained by using the formula of the straight line passing through the point
[Y-（-1）]=[-（-1）]（X-2））
That is y = x-3
The center of the circle is on the line 2x + y = 0, which is the intersection of the line y = x - 3 and it
Solution equation: y = x-3
Y=-2X
We get x = 1, y = - 2, that is, the center of the circle is (1, - 2)
Using the distance formula r = √ [(x1-x2) ^ 2 + (y1-y2) ^ 2], the distance between point (1,2) and point a (2, - 1) should be √ 2 -------- (^ denotes power)
Using the formula of circle with center a (a, b) and radius R: (x-a) ^ 2 + (y-b) ^ 2 = R ^ 2
Then we get the equation (x-1) ^ 2 + (y + 2) ^ 2 = 2 of circle M