A circle has the equation \( 4(x-2)^{2}+4 y^{2}=4 \). (a) Find the center ( \( h, k \) ) and radius \( r \) of the circle (b) Graph the circle. (c) Find the intercepts, if any, of the graph.

Certainly! Let's break down each part of the problem step by step. --- ## **Problem Statement** A circle has the equation: \[ 4(x - 2)^2 + 4y^2 = 4 \] **Tasks:** a) Find the center \((h, k)\) and radius \(r\) of the circle. b) Graph the circle. c) Find the intercepts, if any, of the graph. --- ## **Part (a): Finding the Center and Radius** To identify the center and radius of the circle, we'll rewrite the given equation in the **standard form** of a circle's equation: \[ (x - h)^2 + (y - k)^2 = r^2 \] where \((h, k)\) is the center and \(r\) is the radius. **Given Equation:** \[ 4(x - 2)^2 + 4y^2 = 4 \] **Steps:** 1. **Divide both sides by 4** to simplify: \[ \frac{4(x - 2)^2}{4} + \frac{4y^2}{4} = \frac{4}{4} \] \[ (x - 2)^2 + y^2 = 1 \] 2. **Compare with the standard form**: \[ (x - h)^2 + (y - k)^2 = r^2 \] From the comparison: - \( h = 2 \) - \( k = 0 \) - \( r^2 = 1 \) ⇒ \( r = 1 \) **Answer:** - **Center \((h, k)\):** \((2, 0)\) - **Radius \(r\):** \(1\) --- ## **Part (b): Graphing the Circle** While I can't provide a graphical image directly, I can guide you on how to sketch the circle based on the center and radius found. **Steps to Graph:** 1. **Plot the Center:** - Locate the point \((2, 0)\) on the Cartesian plane. 2. **Draw the Circle:** - Using the radius \(r = 1\), draw a circle that extends 1 unit in all directions from the center. - The circle will touch the points: - **Right:** \((3, 0)\) - **Left:** \((1, 0)\) - **Top:** \((2, 1)\) - **Bottom:** \((2, -1)\) **Visual Guide:** ``` y | 1 | • (2,1) | 0 | • (1,0) • (2,0) • (3,0) | -1| • (2,-1) | +--------------------------- x ``` - The center is at \((2, 0)\). - The circle has a radius of 1, connecting the center to each of the four points shown. --- ## **Part (c): Finding the Intercepts** **Intercepts** are points where the graph crosses the axes. ### **1. X-Intercepts** To find the x-intercepts, set \( y = 0 \) and solve for \( x \). **Given Equation:** \[ (x - 2)^2 + y^2 = 1 \] **Set \( y = 0 \):** \[ (x - 2)^2 + 0^2 = 1 \] \[ (x - 2)^2 = 1 \] \[ x - 2 = \pm 1 \] \[ x = 2 \pm 1 \] \[ x = 3 \quad \text{or} \quad x = 1 \] **X-Intercepts:** - \((3, 0)\) - \((1, 0)\) ### **2. Y-Intercepts** To find the y-intercepts, set \( x = 0 \) and solve for \( y \). **Set \( x = 0 \):** \[ (0 - 2)^2 + y^2 = 1 \] \[ 4 + y^2 = 1 \] \[ y^2 = 1 - 4 \] \[ y^2 = -3 \] Since \( y^2 \) cannot be negative in the set of real numbers, there are **no y-intercepts**. --- ## **Summary of Intercepts** - **X-Intercepts:** \((1, 0)\) and \((3, 0)\) - **Y-Intercepts:** None --- Feel free to ask if you have any further questions or need additional explanations!