(a) Completing both squares, \[(x+3)^2+(y-1)^2=36.\] The centre is \((-3,1)\), the radius is 6, and the lowest point is \((-3,-5)\).

(b) Substitute \(y=kx-5\) into the circle: \[(1+k^2)x^2+(6-12k)x+9=0.\] For two distinct intersections, the discriminant must be positive: \[(6-12k)^2-36(1+k^2)>0\quad\Longrightarrow\quad108k^2-144k>0.\] Hence \[k(3k-4)>0,\qquad k<0\text{ or }k>\frac43.\] At the excluded endpoints the line is tangent.