Question

Find all points on the circle \(x^{2}+y^{2}=100\) where the slope is \(\frac{3}{4} .\)

Short answer

The points on the given circle where the slope of the tangent is 3/4 are (6,-8) and (-6, 8).

Step-by-step solution

Define the Function

The function is defined by the equation of the given circle, which is \(f(x, y) = x^{2} + y^{2} - 100\)

Compute the Derivative

The derivative of the function \(f\) with respect to \(x\) is \[f'(x, y) = 2x + 2y \frac{dy}{dx} = 0\]Solving for \(\frac{dy}{dx}\) gives \[\frac{dy}{dx} = -\frac{x}{y}\]

Equate the Derivative to the Given Slope

The slope of the tangent to the circle at a point is given by the value of \(\frac{dy}{dx}\) at that point. So, set the expression for \(\frac{dy}{dx}\) equal to the given slope 3/4, which gives\[-\frac{x}{y} = \frac{3}{4}\]Cross multiply and solve for \(y\) to obtain \[y=-\frac{4x}{3}\]

Substituting and Finding the Points

Substitute \(y\) in the equation of the circle \(x^{2} + y^{2} = 100\), we obtain the equation\[x^2 + \left(-\frac{4x}{3}\right)^2 = 100\]which simplifies to \[\frac{25x^2}{9} = 100\]Solving for \(x\) gives \(x = 6\) and \(x = -6\). Substituting these \(x\)-values into the equation of the line i.e., \(y=-\frac{4x}{3}\) gives, \(y=-8\) and \(y=8\) when \(x = 6\) and \(x = -6\) respectively. So, the points where the slope of the tangent to the circle is 3/4 are (6, -8) and (-6, 8).

Key concepts

Understanding the Derivative of a Circle

A circle, in its simplest form, is defined by the equation \(x^2 + y^2 = r^2\), where \(r\) is the radius of the circle. To understand how the derivative plays a role, consider the equation you've started with: \(x^2 + y^2 = 100\). Here, if we take the derivative with respect to \(x\), we employ implicit differentiation since \(y\) is also dependent on \(x\). Implicit differentiation helps us find \(\frac{dy}{dx}\).

The derivative essentially measures how the \(y\)-coordinate changes with respect to the \(x\)-coordinate as we move around the circle. Calculating the derivative gives us \(f'(x, y) = 2x + 2y \frac{dy}{dx} = 0\). From here, isolating for \(\frac{dy}{dx}\) gives the expression \(\frac{dy}{dx} = -\frac{x}{y}\). This derivative shows the rate of change of \(y\) with respect to \(x\) for a point on the circle.

Tangent Slope on a Circle

The tangent to a circle at any point is a straight line that just "touches" the circle at that point. The slope of this tangent line is given by the derivative \(\frac{dy}{dx}\) at that specific point. In our exercise, we want this slope to be \(\frac{3}{4}\).

Equating our derivative to the slope, we have \(\frac{dy}{dx} = -\frac{x}{y} = \frac{3}{4}\). Solving this equation helps us find the relationship between \(x\) and \(y\) for the tangent to have the specified slope. This leads us to the line equation \(y = -\frac{4x}{3}\). Understanding this relationship is crucial since it pinpoints where on our circle a tangent with a given slope occurs.

Identifying Coordinate Points on the Circle

To find the points where a tangent with the slope \(\frac{3}{4}\) touches the circle, we substitute the equation derived from the slope back into the original circle equation. This step involves using \(y = -\frac{4x}{3}\) in \(x^2 + y^2 = 100\).

By substitution, we solve the equation \(x^2 + \left(-\frac{4x}{3}\right)^2 = 100\), simplifying to \(\frac{25x^2}{9} = 100\). Solving this quadratic equation gives the \(x\)-coordinates \(x = 6\) and \(x = -6\), and using our line equation \(y = -\frac{4x}{3}\), we find the corresponding \(y\)-coordinates: \(y = -8\) for \(x = 6\) and \(y = 8\) for \(x = -6\). Thus, the points are \((6, -8)\) and \((-6, 8)\).

Solving Equations Using Substitution

Substitution is a powerful technique used to find solutions by replacing one variable with an equivalent expression involving another variable. Here, it simplifies the process of finding specific coordinate points on the circle.

After establishing that \(y = -\frac{4x}{3}\), substitution into the circle's equation, \(x^2 + y^2 = 100\), allows us to express everything in terms of one variable \(x\). This reduces a two-variable problem into a single-variable problem which is often easier to solve.

• First, substitute \(y = -\frac{4x}{3}\) into the circle's equation.
• Then, solve the resulting quadratic equation for \(x\).
• Substitute the solved \(x\)-values back to find the corresponding \(y\)-values.

By understanding and using substitution, you can systematically find the exact points of interest on the circle.