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.

To solve the problem, let's begin with the first part: (a) The given equation of the circle is \( 4(x-2)^{2}+4 y^{2}=4 \). We can simplify this equation by dividing everything by 4: \[ (x-2)^{2} + y^{2} = 1. \] This shows us the standard form of the circle's equation, which is \((x-h)^2 + (y-k)^2 = r^2\). From this, we can identify the center \( (h, k) \) as \( (2, 0) \) and the radius \( r \) as \( 1 \) (since \( r^2 = 1 \)). (b) To graph the circle, plot the center at the point \( (2, 0) \) on the Cartesian plane. From the center, move 1 unit in all four directions: up, down, left, and right. This creates a circle that touches the points \( (1, 0) \), \( (3, 0) \), \( (2, 1) \), and \( (2, -1) \). (c) To find the intercepts, check where the circle intersects the x-axis and y-axis. For the x-intercepts, set \( y = 0 \): \[ (x-2)^{2} + 0^{2} = 1 \implies (x-2)^{2} = 1. \] This gives \( x-2 = 1 \) or \( x-2 = -1 \), yielding \( x = 3 \) and \( x = 1 \). So, the x-intercepts are \( (1, 0) \) and \( (3, 0) \). For the y-intercepts, set \( x = 0 \): \[ (0-2)^{2} + y^{2} = 1 \implies 4 + y^{2} = 1 \implies y^{2} = -3. \] Since \( y^{2} = -3 \) has no real solutions, the circle has no y-intercepts. **Summary:** - Center \( (h, k) = (2, 0) \) - Radius \( r = 1 \) - X-intercepts at \( (1, 0) \) and \( (3, 0) \); no y-intercepts.