Chapter 1: Problem 72
Suppose that \((a, b)\) is on the circle \(x^{2}+y^{2}=r^{2}\). Show that the line \(a x+b y=r^{2}\) is tangent to the circle at \((a, b)\).
Short Answer
Expert verified
The line is tangent as its slope is the negative reciprocal of the radius's slope, showing perpendicularity.
Step by step solution
01
Understand the Circle Equation
The equation of the circle is given by \(x^2 + y^2 = r^2\), with center at \((0, 0)\) and radius \(r\). The point \((a, b)\) lies on the circle, so it satisfies the equation: \(a^2 + b^2 = r^2\).
02
Write the Tangent Line Equation
A line tangent to the circle at \((a, b)\) will have the equation \(ax + by = r^2\). This is derived from the principle that the tangent line to the circle at any point on its boundary forms a right angle with the radius at that point.
03
Calculate the Gradient of the Radius
The gradient (or slope) of the radius from the center \((0, 0)\) to the point \((a, b)\) on the circle is \(\frac{b - 0}{a - 0} = \frac{b}{a}\).
04
Determine the Gradient of the Tangent Line
The gradient of the tangent line \(ax + by = r^2\) can be found by rearranging it to the slope-intercept form \(y = -\frac{a}{b}x + \frac{r^2}{b}\), which gives a slope of \(-\frac{a}{b}\).
05
Confirm Perpendicularity to Verify Tangency
For the line to be tangent, its gradient \(-\frac{a}{b}\) should be the negative reciprocal of the radius's gradient \(\frac{b}{a}\). Calculating \(-\left(\frac{b}{a}\right)\) gives \(-\frac{a}{b}\), confirming that the line at \(ax + by = r^2\) is tangent to the circle at \((a, b)\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Circle Equation
At the heart of many geometry problems is the circle equation, which is a fundamental concept in mathematics. The equation of a circle in its standard form is \(x^2 + y^2 = r^2\). Here, the origin \((0, 0)\) serves as the center of the circle, and \(r\) represents the radius, the distance from the center to any point on the circle's circumference. Any point \((x, y)\) that satisfies this equation lies on the circle.
For example, a point \((a, b)\) is truly on the circle if plugging it into the circle equation holds true: \(a^2 + b^2 = r^2\). Comprehending this equation helps us understand the basic geometric property that all points equidistant from the center form the circle.
Thus, visualizing where the tangent line meets this circle is easier. It ensures we know there's a point \((a, b)\) on its boundary that satisfies the circle's equation.
For example, a point \((a, b)\) is truly on the circle if plugging it into the circle equation holds true: \(a^2 + b^2 = r^2\). Comprehending this equation helps us understand the basic geometric property that all points equidistant from the center form the circle.
Thus, visualizing where the tangent line meets this circle is easier. It ensures we know there's a point \((a, b)\) on its boundary that satisfies the circle's equation.
The Role of the Radius
The radius is pivotal in defining both the shape and the properties of a circle. In geometry, it is a straight line all the way from the center to the circle's edge. The length of this line is known as the radius with each radius in a circle being the same length.
Since the radius connects the center to a point \((a, b)\) on the circle, it plays a significant role in determining various equations, including equations of tangents. From the center \((0, 0)\) to the point \((a, b)\), the length is \(r\), conforming with \(a^2 + b^2 = r^2\), reiterating the invariant nature of the radius.
Furthermore, from any point \((a, b)\) on the circle, the radius forms a right angle with the tangent line at this point. This concept of perpendicularity is crucial in understanding the relationship between a circle and its tangent lines.
Since the radius connects the center to a point \((a, b)\) on the circle, it plays a significant role in determining various equations, including equations of tangents. From the center \((0, 0)\) to the point \((a, b)\), the length is \(r\), conforming with \(a^2 + b^2 = r^2\), reiterating the invariant nature of the radius.
Furthermore, from any point \((a, b)\) on the circle, the radius forms a right angle with the tangent line at this point. This concept of perpendicularity is crucial in understanding the relationship between a circle and its tangent lines.
Gradient in Relation to the Circle
The gradient, also known as the slope, tells us how steep a line is. When working with circles, it is essential to calculate the gradients of different lines, such as the radius or the tangent.
The radius line's gradient from the center \((0, 0)\) to \((a, b)\) is found as \(\frac{b}{a}\). This indicates the steepness of the line from the center to the point on the circle. Understanding this gradient helps when determining the tangent line's equation.
Conversely, for the tangent line \(ax + by = r^2\), rewriting it into the slope-intercept form \(y = -\frac{a}{b}x + \frac{r^2}{b}\) brings out its gradient as \(-\frac{a}{b}\). Recognizing these gradients is key when we use perpendicularity principles to confirm that the tangent's gradient is the negative reciprocal of the radius's gradient.
The radius line's gradient from the center \((0, 0)\) to \((a, b)\) is found as \(\frac{b}{a}\). This indicates the steepness of the line from the center to the point on the circle. Understanding this gradient helps when determining the tangent line's equation.
Conversely, for the tangent line \(ax + by = r^2\), rewriting it into the slope-intercept form \(y = -\frac{a}{b}x + \frac{r^2}{b}\) brings out its gradient as \(-\frac{a}{b}\). Recognizing these gradients is key when we use perpendicularity principles to confirm that the tangent's gradient is the negative reciprocal of the radius's gradient.
The Concept of Perpendicularity
Perpendicularity is an essential concept when discussing the relationship between a circle and its tangent. Perpendicular lines meet at a right angle (90 degrees), a crucial aspect of understanding tangency.
For a line to be tangent, it must be perpendicular to the radius at the point of tangency. You can check this by confirming that the product of their gradients is \(-1\). In our example, the radius' gradient is \(\frac{b}{a}\), and the tangent's gradient is \(-\frac{a}{b}\). Multiplying these together: \(\frac{b}{a} \times -\frac{a}{b} = -1\), showing clear perpendicularity.
This verification of perpendicularity confirms that the line \(ax + by = r^2\) is indeed tangent to the circle at the point \((a, b)\). Understanding this relationship is critical in many geometric proofs and applications.
For a line to be tangent, it must be perpendicular to the radius at the point of tangency. You can check this by confirming that the product of their gradients is \(-1\). In our example, the radius' gradient is \(\frac{b}{a}\), and the tangent's gradient is \(-\frac{a}{b}\). Multiplying these together: \(\frac{b}{a} \times -\frac{a}{b} = -1\), showing clear perpendicularity.
This verification of perpendicularity confirms that the line \(ax + by = r^2\) is indeed tangent to the circle at the point \((a, b)\). Understanding this relationship is critical in many geometric proofs and applications.
