Answer
Equation of tangent:
$$3x+y-10=0$$
Equation of Normal:
$$x-3y=0$$
Work Step by Step
Step-1: We need to find slope of the curve at $(3,1)$. So, we will differentiate $x^2 + y^2 = 10$ with respect to $x$,
$$2x + 2y\frac{dy}{dx} = 0$$
$$\implies \frac{dy}{dx} = -\frac{x}{y}$$
At $(3, 1)$, slope $m = \frac{dy}{dx} = -3$.
Step-2: Using point-slope form, we determine the equation of the tangent,
$$y-1 = -3(x-3)\implies3x+y-10=0$$
Step-3: Let the slope of the normal be $m_n$. Thus, $m_n \times m = -1\implies m_n = 1/3$. Using the slope-point form, we will determine the equation of normal.
$$y-1=\frac{1}{3}(x-3)\implies x-3y=0$$
Step-4: Picture of the graph is given below.
