Question

write the y - intercept of the equation of the tangent line to $x^{2}+y^{2}=25$ in the third quadrant where $x = - 3$.

Explanation:

Step1: Find the y - value

Given $x^{2}+y^{2}=25$ and $x = - 3$. Substitute $x=-3$ into the equation:
$(-3)^{2}+y^{2}=25$, which simplifies to $9 + y^{2}=25$, then $y^{2}=16$, so $y=\pm4$. Since the point is in the third - quadrant, $y=-4$.

Step2: Differentiate the equation implicitly

Differentiate $x^{2}+y^{2}=25$ with respect to $x$.
Using the chain - rule, we have $2x + 2y\frac{dy}{dx}=0$.
Solve for $\frac{dy}{dx}$: $\frac{dy}{dx}=-\frac{x}{y}$.

Step3: Find the slope of the tangent line

Substitute $x=-3$ and $y = - 4$ into $\frac{dy}{dx}$:
$\frac{dy}{dx}=-\frac{-3}{-4}=-\frac{3}{4}$.

Step4: Write the equation of the tangent line

Use the point - slope form $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(-3,-4)$ and $m = -\frac{3}{4}$.
$y+4=-\frac{3}{4}(x + 3)$.

Step5: Find the y - intercept

Set $x = 0$ in the equation of the tangent line:
$y+4=-\frac{3}{4}(0 + 3)$,
$y+4=-\frac{9}{4}$,
$y=-\frac{9}{4}-4=-\frac{9 + 16}{4}=-\frac{25}{4}$.

Answer:

$-\frac{25}{4}$