Answer:
[tex]\displaystyle y= -\frac{3}{4} x + \frac{25}{4}[/tex].
Step-by-step explanation:
The equation of a circle of radius [tex]5[/tex] centered at [tex](0,0)[/tex] is:
[tex]x^{2} + y^{2} = 5^{2}[/tex].
[tex]x^{2} + y^{2} = 25[/tex].
Differentiate implicitly with respect to [tex]x[/tex] to find the slope of tangents to this circle.
[tex]\displaystyle \frac{d}{dx}[x^{2} + y^{2}] = \frac{d}{dx}[25][/tex]
[tex]\displaystyle \frac{d}{dx}(x^{2}) + \frac{d}{dx}(y^{2}) = 0[/tex].
Apply the power rule and the chain rule. Treat [tex]y[/tex] as a function of [tex]x[/tex], [tex]f(x)[/tex].
[tex]\displaystyle \frac{d}{dx}(x^{2}) + \frac{d}{dx}(f(x))^{2} = 0[/tex].
[tex]\displaystyle \frac{d}{dx}(2x) + \frac{d}{dx}(2f(x)\cdot f^{\prime}(x)) = 0[/tex].
That is:
[tex]\displaystyle \frac{d}{dx}(2x) + \frac{d}{dx}\left(2y \cdot \frac{dy}{dx}\right) = 0[/tex].
Solve this equation for [tex]\displaystyle \frac{dy}{dx}[/tex]:
[tex]\displaystyle \frac{dy}{dx} = -\frac{x}{y}[/tex].
The slope of the tangent to this circle at point [tex](3, 4)[/tex] will thus equal
[tex]\displaystyle \frac{dy}{dx} = -\frac{3}{4}[/tex].
Apply the slope-point of a line in a cartesian plane:
[tex]y - y_0 = m(x - x_0)[/tex], where
- [tex]m[/tex] is the gradient of this line, and
- [tex](x_0, y_0)[/tex] are the coordinates of a point on that line.
For the tangent line in this question:
- [tex]\displaystyle m = -\frac{3}{4}[/tex],
- [tex](x_0, y_0) = (3, 4)[/tex].
The equation of this tangent line will thus be:
[tex]\displaystyle y - 4 = -\frac{3}{4} (x - 3)[/tex].
That simplifies to
[tex]\displaystyle y= -\frac{3}{4} x + \frac{25}{4}[/tex].
