Given that the line x + 2y-3 = 0, the intersection circle x ^ 2 + y ^ 2 + x-6y + F = 0 is at the point P, Q, O, and OP is perpendicular to OQ, then f is?

Given that the line x + 2y-3 = 0, the intersection circle x ^ 2 + y ^ 2 + x-6y + F = 0 is at the point P, Q, O, and OP is perpendicular to OQ, then f is?

Is the circular equation x ^ 2 + y ^ 2 + x-6y + F = 0 an extra X?
Otherwise, the calculation will be troublesome
Connecting PQ
Do od vertical PQ to d
Because OP = OQ
The straight line od is the bisector of the angle ∠ poq
∠POQ = 90°
∠POD = 45°
r = OP = √2OD
Od is the distance from the center of the circle to the line x + 2y-3 = 0
The circle: x ^ 2 + y ^ 2-6y + F = 0 is changed into the standard form
x^2 + (y-3)^2 = 9 - F
Then the center of the circle is (0,3)
Radius r = √ (9 - F)
OD = |6-3|/√5 = 3√5/5
So there are
√(9 - F) = √2*3√5/5
Solution
F = 27/5
If the circular equation is correct, the calculation method is still the same, but the calculation is much more complicated

Given that the intersection points of circle X & # 178; + Y & # 178; + x-6y + C = 0 and straight line x2y-3 = 0 are p and Q, and op ⊥ OQ (o is the origin of coordinates), the equation of circle is solved
The linear equation is x + 2y-3 = 0. I'm sorry, I'm in a hurry,

If you don't write your linear equation clearly, please add it first

If the system of inequalities x-m > n x + M

From x-m > n x + M