Answer:
(2,-3) and (49/29, -108/29)
Step-by-step explanation: Assuming this is an exam and time is limited, first you gotta sketch the circle for clarity. Using the standard circle equation
(x-a)^2 + (y-b)^2 = r^2
(x-1)^2 + (y+3)^2 =1
You can always expand to verify it gives x^2 + y^2 - 2x + 6y + 9=0
Now you know (a,b) which is also the center of the circle being= (1,-3) and the radius(r) = 1
Your first assumption here even without drawing the circle is that it has to be in the fourth quadrant meaning all values of x for any point of intersection on that circle has to be +ve. Looking at your answer choices only c with is (2,-3) and (49/29, -108/29) fulfills this hence it is your answer.
Also you have to realize point (2,-3) on the straight line has same y coordinates with center of this circle.