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(e) graph ((x - 4)^2 + (y + 3)^2 = 49)

Question

(e) graph ((x - 4)^2 + (y + 3)^2 = 49)

Explanation:

Step1: Recall circle equation form

The standard form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius.

Step2: Identify center and radius

For the equation \((x - 4)^2 + (y + 3)^2 = 49\), rewrite \(y + 3\) as \(y - (-3)\). So, \(h = 4\), \(k = -3\), and \(r^2 = 49\), which means \(r = \sqrt{49}=7\).

Step3: Plot the center

Locate the point \((4, -3)\) on the coordinate plane.

Step4: Draw the circle

With the center at \((4, -3)\) and radius \(7\), draw a circle. The radius means the distance from the center to any point on the circle is \(7\) units. For example, moving \(7\) units right from the center: \(x = 4 + 7 = 11\) (but the grid ends at \(10\), so we can see part), \(7\) units left: \(x = 4 - 7 = -3\), \(7\) units up: \(y = -3 + 7 = 4\), \(7\) units down: \(y = -3 - 7 = -10\) (also outside the visible grid, but the center and the general shape can be plotted with the given grid).

Answer:

To graph \((x - 4)^2 + (y + 3)^2 = 49\):

  1. Center: \((4, -3)\) (from the standard circle equation \((x - h)^2 + (y - k)^2 = r^2\), where \(h = 4\), \(k = -3\)).
  2. Radius: \(r = 7\) (since \(r^2 = 49\), so \(r=\sqrt{49}=7\)).
  3. Plot the center \((4, -3)\) on the coordinate plane. Then, draw a circle with radius \(7\) (distance from center to any point on the circle is \(7\) units) around this center.