The title gives a general equation of circle X 2+Y 2+X-6Y+3 = 0 ①.
The two points P and Q on the curve of a circle are symmetrical about the straight line kx-y+4=0.
∴ The straight line kx-y+4=0 passes through the center of the circle (-1/2,3).
∴k=2
Let the equation of PQ be: y =- 1/2x+B2.
① ② Simultaneous 5/4 x 2+(4-b) x+b 2-6 b+3 = 0.
Let P(x 1, y 1) and Q(x2, y2).
∵OP⊥OQ
∴OP vector OQ vector =0
∴x 1x2+y 1y2=0
∴x 1x2+(- 1/2x 1+b)(- 1/2x2+b)=0
x 1x 2+ 1/4 x 1x2- 1/2b(x 1+x2)+b^2=0
∫x 1x 2 =(B2-6 b+3)/(5/4)
x 1+x2=(b-4)/(5/4)
Substitution and simplification result in 8B 2-22b+ 15 = 0.
B=3/2 or 5/4.
The equation of PQ is x+2y-3=0 or 2x+4y-5=0.