Q. 38 - Geometry

Question

 In Exercises 35-38 find an equation of the circle described and sketch the graph.

 The circle has center (p, q) and is tangent to the x-axis.

Step-by-Step Solution

Answer

The equation of the circle is ${\left(x-p\right)}^2+{\left(y-q\right)}^2=q^2$

The graph is:

Step-1 – Given

Given that the circle has center (p, q) and is tangent to the x-axis. 

Step-2 – To determine

We have to find the equation of the circle and sketch the graph.

Step-3 – Calculation

We first find the length of the radius:

Let, S(x, y) is the tangent point on the circle.

Since the tangent is the x-axis so on x-axis y = 0.

It means, S(x, 0) = S(p, 0). 

Then we find the radius by finding the distance between (p, q) and (p, 0).

$\begin{array}{l}r=\sqrt{{\left(p-p\right)}^2+{\left(q-0\right)}^2}\\ r=\sqrt{{\left(0\right)}^2+{\left(q\right)}^2}\\ r=\sqrt{{\left(q\right)}^2}\\ r=q\end{array}$

It means, radius = r = q.

 

Here, center = (a, b) = (p, q) and radius = r = q.

We plug them in the standard form of the equation of a circle:

${\left(x-a\right)}^2+{\left(y-b\right)}^2=r^2$

${\left(x-p\right)}^2+{\left(y-q\right)}^2={\left(q\right)}^2$

${\left(x-p\right)}^2+{\left(y-q\right)}^2=q^2$

So, the equation of the circle is${\left(x-p\right)}^2+{\left(y-q\right)}^2=q^2$

Step-4 – Graph

We will sketch the graph using a graphing utility.

Step 1: Press WINDOW button in order to access the Window editor.

Step 2: Press$\boxed{\text{Y=}}$ button.

Step 3: Enter the expression ${\left(x-p\right)}^2+{\left(y-q\right)}^2=q^2$  . (used p = 4 and q = 3).

Step 4: Press GRAPH button to graph the function and then adjust the window.

The obtained graph is:

From the graph, we see that the center is (p, q) and the radius is q.