Given that the circle x + y + x-6y + M = 0 and the line x + 2y-3 = 0 intersect at P and Q, whether there is a real number m, such that PQ is the diameter

Given that the circle x + y + x-6y + M = 0 and the line x + 2y-3 = 0 intersect at P and Q, whether there is a real number m, such that PQ is the diameter

For your idea, the equation of circle can be written in another form, that is, you can see the shape of the center of the circle. You can easily use the coordinates of the center of the circle to solve the linear equation. At this time, there should be only m, an unknown number. If you can solve a real number, it means there is no solution, it means there is no solution

(1 / 2) P is a moving point on the circle x ^ 2 + y ^ 2 + 4x-6y = 0, find (2) the fixed point n (- 2,2), find the maximum value of | PQ | and (3) the point P with the largest distance from the straight line y = 2
(1 / 2) P is a moving point on the circle x ^ 2 + y ^ 2 + 4x-6y = 0, find (2) the fixed point n (- 2,2), find the maximum value of | PQ | and (3) the coordinates of the point P with the largest distance from the straight line y = 2（

The standard equation of circle is (x + 2) ^ 2 + (Y-3) ^ 2 = 13
The distance between Q and circle center is √ [(1 + 2) ^ 2 + (- 1-3) ^ 2] = 5
The minimum value of PQ = distance between two points minus radius = 5 - √ 13
The maximum value of PQ = distance between two points plus radius = 5 + √ 13

Given that the two intersections of circle x ^ 2 + y ^ 2 + x-6x = 0 and straight line x + 2y-3 = 0 are P.Q, the equation of circle with PQ as diameter is obtained

There is something wrong with the title
I'll give you some ideas
Firstly, the coordinates, i.e. P and Q coordinates, are obtained according to the circular equation and linear equation
Then calculate the coordinates of the center point of PQ
Then get the radius
Finally, the circular equation is obtained