/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Suppose that the point \(\left(x... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that the point \(\left(x_{0}, y_{0}\right)\) lies on the circle \(x^{2}+y^{2}=a^{2}\) Show that the equation of the line tangent to the circle at \(\left(x_{0}, y_{0}\right)\) is \(x_{0} x+y_{0} y=a^{2}\).

Short Answer

Expert verified
The tangent line equation is \(x_0x + y_0y = a^2\).

Step by step solution

01

Equation of the Circle

The equation of the circle is given as \(x^2 + y^2 = a^2\). This equation represents all the points \((x, y)\) that are equidistant from the origin with radius \(a\).
02

Identify the Tangent Point

The point \((x_0, y_0)\) lies on the circle, which means it satisfies the circle's equation. Therefore, \(x_0^2 + y_0^2 = a^2\).
03

Use Implicit Differentiation

To find the slope of the tangent line at point \((x_0, y_0)\), differentiate the equation \(x^2 + y^2 = a^2\) implicitly with respect to \(x\).Differentiate: \[\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(a^2)\]Result: \[2x + 2y\frac{dy}{dx} = 0\]
04

Solve for the Derivative

Rearrange the differentiated equation to solve for \(\frac{dy}{dx}\) which represents the slope of the tangent line.\[2x + 2y\frac{dy}{dx} = 0\]\\[2y\frac{dy}{dx} = -2x\]\\[\frac{dy}{dx} = -\frac{x}{y}\]
05

Find the Tangent Line Equation

The slope \(\frac{dy}{dx} = -\frac{x_0}{y_0}\) at the specific point \((x_0, y_0)\). Utilize the point-slope form of the line equation: \(y - y_0 = m(x - x_0)\) where \(m = -\frac{x_0}{y_0}\).
06

Simplify and Rearrange

Substitute \(m\) and \(x_0, y_0\):\[y - y_0 = -\frac{x_0}{y_0}(x - x_0)\] Multiply through by \(y_0\) to eliminate the fraction:\[y_0(y - y_0) = -x_0(x - x_0)\]Simplify:\[y_0y - y_0^2 = -x_0x + x_0^2\] Rearrange terms:\[x_0x + y_0y = x_0^2 + y_0^2\]Since \(x_0^2 + y_0^2 = a^2\), we substitute back to get:\[x_0x + y_0y = a^2\]
07

Conclusion

Thus, the equation for the tangent line at point \((x_0, y_0)\) on the circle is indeed \(x_0x + y_0y = a^2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Implicit Differentiation
Implicit differentiation is a powerful tool used to find derivatives when variables are intermixed in an equation, as opposed to being isolated on one side. In the context of our problem, we have the circle's equation, \(x^2 + y^2 = a^2\). Here, \(y\) is not a direct function of \(x\), making typical differentiation more challenging.
To use implicit differentiation, we differentiate both sides of the equation with respect to \(x\). When differentiating \(x^2\), we proceed as usual because \(x\) is our variable of differentiation. However, for \(y^2\), since \(y\) is a function of \(x\), we apply the chain rule, giving us \(2y\frac{dy}{dx}\).
This method allows us to find \(\frac{dy}{dx}\), the slope of the tangent line, even when our function is not explicitly solved for \(y\) in terms of \(x\). Implicit differentiation is essential in geometry when dealing with curves like circles that are not easily described by simple functions.
Circle
A circle is a fundamental shape in geometry, defined as the set of all points in a plane that are equidistant from a fixed point, called the center. The equation \(x^2 + y^2 = a^2\) describes a circle with its center at the origin \((0, 0)\) and radius \(a\). Each point \((x, y)\) on this circle satisfies the equation, forming the circular shape.
Circles are a common subject in problems involving tangent lines, which are lines that just touch the circle at one point and are perpendicular to the radius at that point. This characteristic makes circles an excellent source of examples for understanding concepts like derivatives in geometry.
Geometry
Geometry is the branch of mathematics that deals with shapes, sizes, and properties of space. Tangent lines in geometry have a special property: they "touch" a curve at only one point without crossing it. In our exercise, the goal was to find the tangent line to the circle at the point \((x_0, y_0)\).
The tangent line to a circle at a given point is perpendicular to the radius drawn to that point. This fact can be demonstrated with calculus, particularly through the use of derivatives, as we find the slope of the line perpendicular to the radius. Understanding these concepts is key to solving geometry problems involving curves.
Derivative Calculation
Calculating derivatives is the process of finding the slope of a function at a given point. For the circle's equation \(x^2 + y^2 = a^2\), implicit differentiation is used to find the derivative \(\frac{dy}{dx}\).
The derivative \(\frac{dy}{dx} = -\frac{x}{y}\) gives the slope of the tangent line to the circle. This slope is crucial because it helps us form the equation of the tangent line at any given point on the circle, such as \((x_0, y_0)\).
By knowing the slope and using the point-slope form of the line equation, we can derive the specific equation of the tangent, confirming that it aligns with the given condition \(x_0x + y_0y = a^2\). Accurate derivative calculation, therefore, plays a vital role in connecting algebraic manipulation to geometric interpretation.

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