Problem 77 - Thomas Calculus

Question

Tangents Find equations for the tangents to the circle \((x-2)^{2}+(y-1)^{2}=5\) at the points where the circle crosses the coordinate axes.

Step-by-Step Solution

Answer

The tangent equations are: y = -2x, y = 2x - 8, y = 2x + 2.

Step 1: Identify the Circle

The given circle is \[(x-2)^{2}+(y-1)^{2}=5.\]This is a circle with center at \((2, 1)\) and radius \(\sqrt{5}.\)

Step 2: Find Points of Intersection with Axes

Find where the circle crosses the x-axis by setting \(y = 0\) and solving \((x-2)^2 + (0-1)^2 = 5.\)This simplifies to \((x-2)^2 + 1 = 5,\) giving \((x-2)^2 = 4.\) The solutions are \(x = 4\) and \(x = 0.\)Thus, the points of intersection on the x-axis are \((4, 0)\) and \((0, 0)\).Next, find where the circle crosses the y-axis by setting \(x = 0\) and solving \((0-2)^2 + (y-1)^2 = 5.\)This simplifies to \(4 + (y-1)^2 = 5,\) giving \((y-1)^2 = 1.\) The solutions are \(y = 2\) and \(y = 0.\)Thus, the points of intersection on the y-axis are \((0, 2)\) and \((0, 0)\).

Step 3: Find the Tangents at the Intersection Points

Using the point (0,0) first:The slope of the radius at \((0, 0)\) is the slope from \((2, 1)\) to \((0, 0)\), given by \(m = \frac{1-0}{2-0} = \frac{1}{2}.\)The slope of the tangent is the negative reciprocal: \(-2.\)Thus, the equation of the tangent at \((0,0)\) is \(y - 0 = -2(x - 0)\) or \(y = -2x.\)At point (4,0):The slope of the radius from \((2, 1)\) to \((4, 0)\) is \(m = \frac{0-1}{4-2} = -\frac{1}{2}.\)The slope of the tangent is \(2.\)Thus, the equation of the tangent is \(y - 0 = 2(x - 4)\) or \(y = 2x - 8.\)At point (0, 2):The slope of the radius from \((2, 1)\) to \((0, 2)\) is \(m = \frac{2-1}{0-2} = -\frac{1}{2}.\)The slope of the tangent is \(2.\)Thus, the equation of the tangent at \((0, 2)\) is \((y - 2) = 2(x - 0)\) or \(y = 2x + 2.\)

Key Concepts

Circle Equation

A circle's equation provides crucial information about its size, shape, and location within a coordinate system. When dealing with circle equations, you often encounter a form like this: \[(x - h)^2 + (y - k)^2 = r^2,\]where:

• \((h, k)\) represents the center of the circle.
• \(r\) signifies the radius.

In our exercise, the circle's equation is \((x-2)^2+(y-1)^2=5\). This tells us:

• The center is \((2, 1)\).
• The radius is \(\sqrt{5}\).

Identifying these components helps you determine intersections and tangents, both of which are important in geometric problem-solving. The standard form of the circle equation is essential in understanding the circle's relationship to various geometric elements, such as axes and tangents.

Coordinate Geometry

Understanding coordinate geometry is key to solving problems involving circles on the plane. It combines algebra and geometry, allowing us to analyze geometric shapes through algebraic equations. Here's why it's important:

• It enables the calculation of distances between points, which is essential for finding radii and tangents.
• By using formulas, you can find slopes and apply them to determine tangent lines.
• It handles different equation forms, like lines and circles, within the coordinate plane.

For example, when identifying points where the circle intersects the axes, coordinate geometry simplifies solving for \(x\) and \(y\). If you consider setting \(y=0\) to find where the circle hits the x-axis, the circle's equation transforms into one you can solve algebraically, revealing intersection points. This technique applies similarly to the y-axis by setting \(x=0\). Through coordinate geometry, we simplify these calculations and properly interpret geometric characteristics of a circle.

Intersection Points

Finding intersection points between a circle and the coordinate axes involves solving the circle's equation by setting one variable at a time to zero. This approach gives specific points where the circle intersects the horizontal or vertical lines of the axes.Start with the x-axis intersection:

• Set \(y = 0\) in the circle equation.
• You get \((x-2)^2+1=5\), which simplifies to \((x-2)^2=4\).
• Solving gives \(x=4\) or \(x=0\), resulting in intersection points \((4, 0)\) and \((0, 0)\).

Next, determine the y-axis intersection:

• Set \(x = 0\) in the equation.
• This becomes \(4+(y-1)^2=5\), leading to \((y-1)^2=1\).
• Solving results in \(y=2\) or \(y=0\), giving points \((0, 2)\) and \((0, 0)\).

Intersection points are not just about locating where the circle meets lines; they are pivotal in calculating tangent lines. Each intersection provides a new point through which tangent lines can be determined, thus enhancing the understanding of the circle's geometric properties. The intersections serve as critical anchors for studying the behavior of lines in proximity to circle structures.