Answer :
To find the equation of the tangent line to the curve [tex]\(x^2 + y^2 = 169\)[/tex] at the point [tex]\((5, -12)\)[/tex], we can follow these steps:
1. Identify the Curve: The equation [tex]\(x^2 + y^2 = 169\)[/tex] represents a circle with a center at the origin [tex]\((0, 0)\)[/tex] and a radius of [tex]\(13\)[/tex] (since [tex]\(13^2 = 169\)[/tex]).
2. Use Implicit Differentiation: To find the slope of the tangent line at any point on the circle, we differentiate the equation implicitly with respect to [tex]\(x\)[/tex].
[tex]\[
\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(169)
\][/tex]
Differentiating, we get:
[tex]\[
2x + 2y \frac{dy}{dx} = 0
\][/tex]
3. Solve for [tex]\(\frac{dy}{dx}\)[/tex]: Rearrange the equation to solve for [tex]\(\frac{dy}{dx}\)[/tex], which represents the slope of the tangent line.
[tex]\[
2y \frac{dy}{dx} = -2x
\][/tex]
[tex]\[
\frac{dy}{dx} = -\frac{x}{y}
\][/tex]
4. Substitute the Given Point [tex]\((5, -12)\)[/tex]: Substitute [tex]\(x = 5\)[/tex] and [tex]\(y = -12\)[/tex] into the slope formula to get the slope at that specific point.
[tex]\[
\frac{dy}{dx} = -\frac{5}{-12} = \frac{5}{12}
\][/tex]
5. Use the Point-Slope Form: The point-slope form of the equation of a line is [tex]\(y - y_1 = m(x - x_1)\)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is the given point. Plug in the slope and point:
[tex]\[
y + 12 = \frac{5}{12}(x - 5)
\][/tex]
6. Rearrange to Standard Form: Rearrange the equation into the standard form [tex]\(Ax + By = C\)[/tex].
Start by multiplying everything by 12 to eliminate the fraction:
[tex]\[
12(y + 12) = 5(x - 5)
\][/tex]
[tex]\[
12y + 144 = 5x - 25
\][/tex]
Rearrange to get:
[tex]\[
5x - 12y = 169
\][/tex]
7. Identify the Correct Answer: Compare this equation to the given options. The correct option is:
[tex]\[
\boxed{C}
\][/tex]
So, the equation of the tangent line to the curve at the point [tex]\((5, -12)\)[/tex] is [tex]\(5x - 12y = 169\)[/tex].
1. Identify the Curve: The equation [tex]\(x^2 + y^2 = 169\)[/tex] represents a circle with a center at the origin [tex]\((0, 0)\)[/tex] and a radius of [tex]\(13\)[/tex] (since [tex]\(13^2 = 169\)[/tex]).
2. Use Implicit Differentiation: To find the slope of the tangent line at any point on the circle, we differentiate the equation implicitly with respect to [tex]\(x\)[/tex].
[tex]\[
\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(169)
\][/tex]
Differentiating, we get:
[tex]\[
2x + 2y \frac{dy}{dx} = 0
\][/tex]
3. Solve for [tex]\(\frac{dy}{dx}\)[/tex]: Rearrange the equation to solve for [tex]\(\frac{dy}{dx}\)[/tex], which represents the slope of the tangent line.
[tex]\[
2y \frac{dy}{dx} = -2x
\][/tex]
[tex]\[
\frac{dy}{dx} = -\frac{x}{y}
\][/tex]
4. Substitute the Given Point [tex]\((5, -12)\)[/tex]: Substitute [tex]\(x = 5\)[/tex] and [tex]\(y = -12\)[/tex] into the slope formula to get the slope at that specific point.
[tex]\[
\frac{dy}{dx} = -\frac{5}{-12} = \frac{5}{12}
\][/tex]
5. Use the Point-Slope Form: The point-slope form of the equation of a line is [tex]\(y - y_1 = m(x - x_1)\)[/tex], where [tex]\(m\)[/tex] is the slope and [tex]\((x_1, y_1)\)[/tex] is the given point. Plug in the slope and point:
[tex]\[
y + 12 = \frac{5}{12}(x - 5)
\][/tex]
6. Rearrange to Standard Form: Rearrange the equation into the standard form [tex]\(Ax + By = C\)[/tex].
Start by multiplying everything by 12 to eliminate the fraction:
[tex]\[
12(y + 12) = 5(x - 5)
\][/tex]
[tex]\[
12y + 144 = 5x - 25
\][/tex]
Rearrange to get:
[tex]\[
5x - 12y = 169
\][/tex]
7. Identify the Correct Answer: Compare this equation to the given options. The correct option is:
[tex]\[
\boxed{C}
\][/tex]
So, the equation of the tangent line to the curve at the point [tex]\((5, -12)\)[/tex] is [tex]\(5x - 12y = 169\)[/tex].
