Use differentiation method to find the slope of the tangent hence theequation of the tangent as shown below.Circle with radius = 5and centre at (-3,1)Tagent of thecircle at x = -6

Answer:The equation of the tangent at x=-6 is [tex]y=-\frac{3}{4}x-\frac{15}{2}[/tex]Step-by-step explanation:The equation of a circle with center (h,k) with radius r units is given by:[tex](x-h)^2+(y-k)^2=r^2[/tex]The given circle has center (-3,1) and radius 5 units.We substitute the center and the radius into the equation to get;[tex](x--3)^2+(y-1)^2=5^2[/tex][tex](x+3)^2+(y-1)^2=25[/tex]To find the slope, we differentiate implicitly to get:[tex]2(x+3)+2(y-1)\fra{dy}{dx}=0[/tex][tex]2(y-1)\frac{dy}{dx}=-2(x+3)[/tex][tex]\frac{dy}{dx}=-\frac{x+3}{y-1}[/tex]When x=-6;we have [tex](-6+3)^2+(y-1)^2=25[/tex] [tex]\implies 9+(y-1)^2=25[/tex] [tex]\implies (y-1)^2=25-9[/tex] [tex]\implies (y-1)^2=16[/tex] [tex]\implies y-1=\pm \sqrt{16}[/tex] [tex]\implies y-1=\pm4[/tex] [tex]\implies y=1\pm4[/tex] [tex]y=-3[/tex] or  [tex]y=5[/tex]From the graph the reuired point is (-6,-3).We substitute this point to find the slope;[tex]\frac{dy}{dx}=-\frac{-6+3}{-3-1}[/tex][tex]\frac{dy}{dx}=-\frac{3}{4}[/tex]The equation is given by [tex]y-y_1=m(x-x_1)[/tex].We plug in the slope and the point to get:[tex]y--3=-\frac{3}{4}(x--6)[/tex][tex]y=-\frac{3}{4}(x+6)-3[/tex][tex]y=-\frac{3}{4}x-\frac{9}{2}-3[/tex][tex]y=-\frac{3}{4}x-\frac{15}{2}[/tex]