91Ó°ÊÓ

To find the x-intercepts of a circle, we need to recognize that they are the points where the circle crosses the x-axis. At these points, the value of y is zero.
Substituting this into the circle's equation helps us determine possible x-values.

The equation of our circle is \[(x - 3)^2 + (y + 6)^2 = 49.\]When we set \(y = 0\), the equation becomes \[(x - 3)^2 + 36 = 49.\]
Next, we isolate the term \((x - 3)^2\) by subtracting 36 from both sides:\[(x - 3)^2 = 13.\]
Upon taking the square root of both sides, we have two potential solutions due to the square root property,\[x - 3 = \pm\sqrt{13}.\]This simplifies to two x-values:

• \(x = 3 + \sqrt{13}\)
• \(x = 3 - \sqrt{13}\)

Therefore, the x-intercepts of the circle are the points \((3 + \sqrt{13}, 0)\) and \((3 - \sqrt{13}, 0)\).
These solutions show where the circle crosses the x-axis. This concept is crucial for graphing as it tells us the extent to which the circle spreads across the x-axis.

Finding the y-intercepts follows a similar process to finding the x-intercepts. While x-intercepts occur at points where the circle touches the x-axis, y-intercepts occur where the circle crosses the y-axis. Thus, the x-value for y-intercepts is zero, or \(x = 0\).

Insert \(x = 0\) into the circle's equation\[(0 - 3)^2 + (y + 6)^2 = 49.\]This simplifies to\[9 + (y + 6)^2 = 49.\]To solve for y, first we subtract 9 from both sides of the equation:\[(y + 6)^2 = 40.\]Next, take the square root of both sides:\[y + 6 = \pm\sqrt{40}.\]Further simplifying \(\sqrt{40}\) gives \(2\sqrt{10}\), and solving for y results in two distinct solutions:

• \(y = -6 + 2\sqrt{10}\)
• \(y = -6 - 2\sqrt{10}\)

Thus, the y-intercepts are at the points:\((0, -6 + 2\sqrt{10})\) and \((0, -6 - 2\sqrt{10})\).
These points illustrate where the circle extends vertically to touch the y-axis.

The circle's equation reveals important characteristics such as the center and radius, which comprehensively describe the circle's position and size.

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

From the equation \[(x - 3)^2 + (y + 6)^2 = 49,\]we identify the circle's center at \((3, -6)\) and square \(49\) to determine the radius because \(r^2 = 49\).

Calculating the radius gives:\[r = \sqrt{49} = 7.\]

• The center \((3, -6)\) tells us the circle is shifted three units to the right and six units down compared to the origin.
• With a radius of 7, the circle extends 7 units outward from the center in all directions, defining its size.

Understanding the center and radius is integral not just for finding intercepts, but also for graphing the circle accurately. This knowledge allows us to precisely sketch the circle's position on a coordinate plane, knowing exactly where its middle lies and how far it stretches.