It is known that the center of circle m is on the straight line 2x-y + 5 = 0 and intersects with y axis at two points a (0, - 2), B (0,4) (1) to solve the equation of circle M (2) Find the tangent equation of circle m passing through point C (- 4,4); (3) given D (1,3), point P moves on circle m, find the trajectory equation of vertex Q of parallelogram adqp with AD and AP as a group of adjacent sides

It is known that the center of circle m is on the straight line 2x-y + 5 = 0 and intersects with y axis at two points a (0, - 2), B (0,4) (1) to solve the equation of circle M (2) Find the tangent equation of circle m passing through point C (- 4,4); (3) given D (1,3), point P moves on circle m, find the trajectory equation of vertex Q of parallelogram adqp with AD and AP as a group of adjacent sides


1) Because the circle intersects with the Y axis at a and B, the center of the circle is on the vertical bisector of AB, that is, the ordinate of the center is (- 2 + 4) / 2 = 1, in 2x-y + 5 = 0, let y = 1 get x = - 2, so the center of the circle is m (- 2,1), R ^ 2 = ma ^ 2 = (- 2-0) ^ 2 + (1 + 2) ^ 2 = 13, so the equation of the circle m is (x + 2) ^ 2 + (Y-1) ^ 2 = 13.2



If the intersection of circle C and Y axis on the straight line 2x-y-7 = 0 and two points a (0, - 4), B (0, - 2), then the equation of circle C is
Find the center of the radius


Let the center of the circle be (a, 2a-7)
The equation is: (x-a) ^ 2 + (y-2a + 7) ^ 2 = R ^ 2
Substituting a and B, we can get the following results respectively
a^2+(3-2a)^2=r^2
a^2+(5-2a)^2=r^2
By subtracting the two formulas: - 16 + 8A = 0
That is, a = 2
That is, R ^ 2 = 2 ^ 2 + (3-4) ^ 2 = 5
So the equation is: (X-2) ^ 2 + (y + 3) ^ 2 = 5