Question

a) \(P(\theta)\) lies at the intersection of the unit circle and the line joining \(\mathrm{A}(5,2)\) to the origin. Use your knowledge of similar triangles and the unit circle to determine the exact coordinates of \(\mathrm{P}(\theta).\) b) Determine the radius of a larger circle with centre at the origin and passing through point A. c) Write the equation for this larger circle.

Short answer

a) \( \left( \frac{5}{\sqrt{29}}, \frac{2}{\sqrt{29}} \right) \) and \( \left( -\frac{5}{\sqrt{29}}, -\frac{2}{\sqrt{29}} \right) \). b) \( \sqrt{29} \). c) \( x^2 + y^2 = 29 \).

Step-by-step solution

- Derive the slope of the line

The line joins point A(5,2) to the origin (0,0). Derive the slope (m) of this line using the formula for slope: \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Plugging in the points: \( m = \frac{2 - 0}{5 - 0} = \frac{2}{5} \).

- Equation of the line

Using the point-slope form of the equation of a line, \( y = mx \), because the line passes through the origin. Thus, the equation of the line is \( y = \frac{2}{5}x \).

- Intersection with the unit circle

The unit circle is given by the equation \( x^2 + y^2 = 1 \). Substitute \( y = \frac{2}{5}x \) into this equation: \[ x^2 + \left(\frac{2}{5}x\right)^2 = 1 \]. Simplify this to get: \[ x^2 + \frac{4}{25}x^2 = 1 \Rightarrow \frac{29}{25}x^2 = 1 \Rightarrow x^2 = \frac{25}{29} \Rightarrow x = \pm \frac{5}{\sqrt{29}} \].

- Solve for y

Using \( y = \frac{2}{5}x \), plug in \( x = \pm \frac{5}{\sqrt{29}} \). When \( x = \frac{5}{\sqrt{29}} \), \( y = \frac{2}{5} * \frac{5}{\sqrt{29}} = \frac{2}{\sqrt{29}} \). When \( x = -\frac{5}{\sqrt{29}} \), \( y = \frac{2}{5} * -\frac{5}{\sqrt{29}} = -\frac{2}{\sqrt{29}} \).

- Coordinates of P(θ)

Thus, the exact coordinates where the line intersects the unit circle are \( P(\theta) = \left( \frac{5}{\sqrt{29}}, \frac{2}{\sqrt{29}} \right) \text{ and } P(\theta) = \left( -\frac{5}{\sqrt{29}}, -\frac{2}{\sqrt{29}} \right)\).

- Calculate the radius of the larger circle

The radius of the larger circle is the distance from the origin to point A(5,2). Using the distance formula: \( r = \sqrt{(5-0)^2 + (2-0)^2} = \sqrt{25+4} = \sqrt{29} \).

- Write the equation of the larger circle

The equation of a circle with center at the origin and radius \( \sqrt{29} \) is given by \( x^2 + y^2 = 29 \).

Key concepts

Unit Circle

The unit circle is a circle with a radius of one unit, centered at the origin of a coordinate plane. It is represented by the equation \( x^2 + y^2 = 1 \). This formula essentially says that any point \((x, y)\) on the circle is at a distance of 1 from the origin \((0, 0)\). The unit circle is fundamental in trigonometry and helps us understand how angles relate to coordinates. For solving the problem, the unit circle was key in finding the intersection points with the line extending from point A to the origin.

Similar Triangles

Similar triangles are triangles that have the same shape but may differ in size. They have identical corresponding angles and their corresponding sides are proportional. In this problem, similar triangles help determine the exact coordinates of \(P(\theta)\) on the unit circle. Because the line passing through A(5, 2) and the origin intersects the unit circle, similar triangles help us scale down the larger triangle formed by point A to match the smaller triangle constrained by the unit circle.

Slope of a Line

The slope of a line indicates its steepness and direction. It's key in determining the equation of a line. For a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\), the slope \(m\) is calculated using the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \]In our exercise, the points \(A(5,2)\) and \((0,0)\) yield the slope \( m = \frac{2 - 0}{5 - 0} = \frac{2}{5} \).

Equation of a Line

The equation of a line can be written in several forms, but for lines passing through the origin, the simplest is the point-slope form: \[ y = mx \]Here, \(m\) is the slope. Given the slope \( \frac{2}{5} \), the line's equation becomes: \[ y = \frac{2}{5}x \]. This equation helps find where the line intersects other curves, such as the unit circle.

Intersection Points

To find where a line intersects the unit circle, we substitute the line's equation into the unit circle's equation. For our exercise, we substitute \( y = \frac{2}{5}x \) into \( x^2 + y^2 = 1 \):\[ x^2 + \frac{4}{25}x^2 = 1 \]Then solve for \(x\):\[ x^2 (1 + \frac{4}{25}) = 1 \Rightarrow \frac{29}{25}x^2 = 1 \Rightarrow x^2 = \frac{25}{29} \Rightarrow x = \frac{5}{\root{29}} \] This yields the coordinates for \(P(\theta)\): \( (\frac{5}{\root{29}}, \frac{2}{\root{29}}) \) and \( (\frac{5}{\root{29}}, -\frac{2}{\root{29}})\).

Distance Formula

The distance formula calculates the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) in a coordinate plane:\[ d = \root{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]Applying this to points \((0, 0)\) and \( (5, 2) \), we find: \[ d = \root{(5-0)^2 + (2-0)^2} = \root{25 + 4} = \root{29} \]. This distance serves as the radius of our larger circle.

Circle Equation

A circle with center at the origin and radius \( r \) has the general equation:\[ x^2 + y^2 = r^2 \]In our scenario, with a radius of \( r = \root{29} \), the equation of the larger circle is: \[ x^2 + y^2 = 29 \]. This describes all points \((x,y)\) that lie on a circle centered at the origin with radius \( \root{29} \).