amaths-circle

If the straight line 5x-y-a=0 touches the circle 3x^2+3y^2-2x+4y+b=0 at the point (c,-1), find the values of a, b and c.
from the question
5c+1-a=0
3c^2+3-2c-4+b=0
Also, the center of the circle is (1/3,-2/3)
The slope of straight line 5x-y-a=0 is 5
The slope of straight line joining (1/3,-2/3) and (c,-1) is
(-1+2/3)/(c-1/3)
=-1/(3c-1)
Since they are perpendicular
-1/(3c-1)*5=-1
5=3c-1
c=2
5c+1-a=0, a=5c+1=11
3c^2+3-2c-4+b=0
b=-(3c^2+3-2c-4)=-(12+3-4-4)=-7