The Y-axis of the circle center on 2x-y-7 = intersects at (0, - 4), (0, - 2), and the circle equation is solved

The Y-axis of the circle center on 2x-y-7 = intersects at (0, - 4), (0, - 2), and the circle equation is solved

Midpoint of intersection with axis (0, - 3)
So the center of the circle passes through the straight line y = - 3 and the straight line 2x-y-7 = 0
Center of circle (2, - 3) obtained by simultaneous solution
Radius = √ (2-0) & sup2; + (- 3 + 4) & sup2; = √ 5
The equation of circle: (X-2) & sup2; + (y + 3) & sup2; = 5

The equation of finding the circle whose center is on the straight line 2x-y-7 = 0 and intersects with y axis at two points a (0, - 4) B (0, - 2)

(x-a)^2+(y-b)^2=r^2;
B = (- 4-2) / 2 = - 3;
2a-b-7=0,a=5;
r^2=(a-0)^2+(b-(-4))^2=26;
So, (X-5) ^ 2 + (y = 3) ^ 2 = 26

The equation of the circle whose center is on the line 2x + y = 0 and tangent to the point (2, - 1) with the line x + Y-1 = 0 is______ ．

Let the center coordinates of the circle be (a, b), then 2A + B = 0|a + B − 1|2 = & nbsp; (a − 2) 2 + (B + 1) 2, the solution is a = 1, B = - 2, so r = 2, so the equation of the circle is (x-1) 2 + (y + 2) 2 = 2

The circle C whose center is on the straight line 2x-y-7 = 0 intersects at two points a (0, - 4) B (0, - 2) on the Y axis to find the equation of circle C,

Let (x-a) ^ 2 + (y-b) ^ 2 = R ^ 2, center O (a, b), radius R
The garden passes a (0, - 4) B (0, - 2) two points
Then a & # 178; + (- 4-b) &# 178; = R & # 178 （1）
a²+(-2-b)²=r²…… （2）
（1）-（2）
B = - 3
If the center of the circle is on the line 2x-y-7 = 0, a = 2
So Yuanxin (2, - 3)
Then R & # 178; = 0A & # 178;
=(0-2)²+(-4+3)²
=4+1
=5
So the equation: (X-2) &# 178; + (y + 3) &# 178; = 5