QuestionIn Exercises $9-12$, find $d y / d x$ and find the slope of the curve at the indicated point.
9. $x^{2}+y^{2}=13, \quad(-2,3)$
10. $x^{2}+y^{2}=9, \quad(0,3)$
11. $(x-1)^{2}+(y-1)^{2}=13, \quad(3,4)$
12. $(x+2)^{2}+(y+3)^{2}=25,(1,-7)$

Studdy Solution
Substitute the point $(-2, 3)$ into the derivative:
$$\frac{dy}{dx} = -\frac{-2}{3} = \frac{2}{3}$$
The slope of the curve at $(-2, 3)$ is $\frac{2}{3}$.
**Exercise 10:**
STEP_1: Differentiate the equation $x^2 + y^2 = 9$ implicitly with respect to $x$:
$$2x + 2y \frac{dy}{dx} = 0$$
STEP_2: Solve for $\frac{dy}{dx}$:
$$\frac{dy}{dx} = -\frac{x}{y}$$
STEP_3: Substitute the point $(0, 3)$ into the derivative:
$$\frac{dy}{dx} = -\frac{0}{3} = 0$$
The slope of the curve at $(0, 3)$ is $0$.
**Exercise 11:**
STEP_1: Differentiate the equation $(x-1)^2 + (y-1)^2 = 13$ implicitly with respect to $x$:
$$2(x-1) + 2(y-1) \frac{dy}{dx} = 0$$
STEP_2: Solve for $\frac{dy}{dx}$:
$$2(y-1) \frac{dy}{dx} = -2(x-1)$$
$$\frac{dy}{dx} = -\frac{x-1}{y-1}$$
STEP_3: Substitute the point $(3, 4)$ into the derivative:
$$\frac{dy}{dx} = -\frac{3-1}{4-1} = -\frac{2}{3}$$
The slope of the curve at $(3, 4)$ is $-\frac{2}{3}$.
**Exercise 12:**
STEP_1: Differentiate the equation $(x+2)^2 + (y+3)^2 = 25$ implicitly with respect to $x$:
$$2(x+2) + 2(y+3) \frac{dy}{dx} = 0$$
STEP_2: Solve for $\frac{dy}{dx}$:
$$2(y+3) \frac{dy}{dx} = -2(x+2)$$
$$\frac{dy}{dx} = -\frac{x+2}{y+3}$$
STEP_3: Substitute the point $(1, -7)$ into the derivative:
$$\frac{dy}{dx} = -\frac{1+2}{-7+3} = -\frac{3}{-4} = \frac{3}{4}$$
The slope of the curve at $(1, -7)$ is $\frac{3}{4}$.