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]
