Question

Use differentiation method to find the slope of the tangent hence the

Answer 1

Answer:

The equation of the tangent at x=-6 is $y=-\frac{3}{4}x-\frac{15}{2}$

Step-by-step explanation:

The equation of a circle with center (h,k) with radius r units is given by:

$(x-h)^2+(y-k)^2=r^2$

The given circle has center (-3,1) and radius 5 units.

We substitute the center and the radius into the equation to get;

$(x--3)^2+(y-1)^2=5^2$

$(x+3)^2+(y-1)^2=25$

To find the slope, we differentiate implicitly to get:

$2(x+3)+2(y-1)\fra{dy}{dx}=0$

$2(y-1)\frac{dy}{dx}=-2(x+3)$

$\frac{dy}{dx}=-\frac{x+3}{y-1}$

When x=-6;we have $(-6+3)^2+(y-1)^2=25$

$\implies 9+(y-1)^2=25$

$\implies (y-1)^2=25-9$

$\implies (y-1)^2=16$

$\implies y-1=\pm \sqrt{16}$

$\implies y-1=\pm4$

$\implies y=1\pm4$

$y=-3$ or  $y=5$

From the graph the reuired point is (-6,-3).

We substitute this point to find the slope;

$\frac{dy}{dx}=-\frac{-6+3}{-3-1}$

$\frac{dy}{dx}=-\frac{3}{4}$

The equation is given by $y-y_1=m(x-x_1)$.

We plug in the slope and the point to get:

$y--3=-\frac{3}{4}(x--6)$

$y=-\frac{3}{4}(x+6)-3$

$y=-\frac{3}{4}x-\frac{9}{2}-3$

$y=-\frac{3}{4}x-\frac{15}{2}$