Problem 36 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the \(x\) - and \(y\) -intercepts of the graph of the circle. $$(x-1)^{2}+(y+4)^{2}=16$$

Step-by-Step Solution

Answer

The x-intercepts are at \(x = 1\) and the y-intercepts are at \(y = \sqrt{15} - 4\) and \(y = -\sqrt{15} - 4\).

Step 1: Substituting y = 0 to find the x-intercepts

We substitute \(y = 0\) in \((x-1)^{2}+(y+4)^{2}=16\) and solve for \(x\). It simplifies to \((x - 1)^{2} + 16 = 16\), which solves down to \(x - 1 = \pm \sqrt{16 - 16} = 0\). Therefore the values of \(x\) are \(x = 1\).

Step 2: Substituting x = 0 to find the y-intercepts

We substitute \(x = 0\) in \((x-1)^{2}+(y+4)^{2}=16\) and solve for \(y\). It simplifies to \((1)^{2} + (y + 4)^{2} = 16\), which solves down to \(y + 4 = \pm \sqrt{16 - 1} = \pm \sqrt{15}\). Therefore the values of \(y\) are \(y = \pm \sqrt{15} - 4\).

Key Concepts

Understanding x-intercepts

To find the x-intercepts of a circle's graph, we set the y variable to zero. This is because x-intercepts are the points where the graph crosses the x-axis, meaning the y-coordinate at these points is zero. In our circle equation, \[(x-1)^{2}+(y+4)^{2}=16\]we replace y with zero in order to find x-intercepts. The equation simplifies to:\[(x - 1)^{2} + 16 = 16\]Solving, we subtract 16 from both sides and arrive at:\[(x - 1)^{2} = 0\]This implies that:\[x - 1 = 0 \ x = 1\]Thus, the x-intercept is at the coordinate (1, 0). This shows that the circle touches the x-axis at this point. Remember, there can be one, two, or no x-intercepts for a circle depending on its position relative to the x-axis.

Unraveling y-intercepts

Y-intercepts occur where a graph crosses the y-axis. Here, the x-coordinate is zero because the graph meets the y-axis directly at these points. We use the circle equation:\[(x-1)^{2}+(y+4)^{2}=16\]Substitute x = 0 into the equation to find the y-intercepts:\[(0 - 1)^{2} + (y + 4)^{2} = 16\]This simplifies to:\[1 + (y + 4)^{2} = 16\]Subtracting 1 from both sides gives:\[(y + 4)^{2} = 15\]Taking the square root results in:\[y + 4 = \, \pm \sqrt{15}\]Solving for y, we find:\[y = -4 \, \pm \sqrt{15}\]This gives us two y-intercepts at

• (0, -4 + \(\sqrt{15}\))
• (0, -4 - \(\sqrt{15}\))

The circle crosses the y-axis at these points, giving us our y-intercepts.

Exploring the equation of a circle

The equation of a circle in standard form is \[(x - h)^{2} + (y - k)^{2} = r^{2}\]where

• \((h, k)\) is the center of the circle.
• \(r\) is the radius.

Our circle's given equation \[(x-1)^{2} + (y+4)^{2} = 16\]matches this standard format. Here, the center is at (1, -4) and the radius squared is 16, which means the radius \(r\) is 4.The center tells us where the circle is positioned on the coordinate plane. It directly impacts whether the circle will intersect the x- or y-axis. A larger radius covers more area, potentially allowing more intercepts with the axes.Understanding this standard circle equation form allows us to easily identify key properties of any circle described by such an equation.