8. Let the equation be A(x+ 1)+B(y-6)=0. Because of tangency, the distance from the center of the circle to the straight line is equal to the radius.
That is | a (-3+1)+b (2-6) |/√ (a2+B2) = 2,
Simplified as B(3B+4A)=0,
Take A= 1, B=0 or A=3, B= -4,
Linear equation can get x+ 1=0 or 3x-4y+27=0.
9. The formula of the circular equation is (x-1) 2+y 2 =1,so the center of the circle is (1,0) and the radius is r= 1.
Because the straight line is tangent to the circle, the distance from the center of the circle to the straight line is equal to the radius of the circle.
That is |1+a+0+1|/√ [(1+a) 2+1] =1,
The solution is a=-1.
10, center (3,0), radius 3,
The distance from that cent of the circle to the straight line is d=|9-0-4|/5= 1,
D 2+(L/2) 2 = R 2 of Pythagorean Theorem,
The solution is L=4√2.