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Q. 36

Expert-verified
Found in: Page 527

### Geometry

Book edition Student Edition
Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen
Pages 227 pages
ISBN 9780395977279

# In Exercises 35-38 find an equation of the circle described and sketch the graph. The circle has center (-2, -4) and passes through point (3, 8).

The equation of the circle is${\left(x+2\right)}^{2}+{\left(y+4\right)}^{2}=169$

The graph is:

See the step by step solution

## Step-1 – Given

Given that the circle has center = (-2, -4) and passes through the point (3, 8).

## 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 by finding the distance between (-2, -4) and (3, 8).

$\begin{array}{l}r=\sqrt{{\left(3-\left(-2\right)\right)}^{2}+{\left(8-\left(-4\right)\right)}^{2}}\\ r=\sqrt{{\left(3+2\right)}^{2}+{\left(8+4\right)}^{2}}\\ r=\sqrt{{\left(5\right)}^{2}+{\left(12\right)}^{2}}\\ r=\sqrt{25+144}\\ r=\sqrt{169}\\ r=13\end{array}$

Here, center = (a, b) = (-2, -4) and radius = r = 13.

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

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

${\left(x+2\right)}^{2}+{\left(y+4\right)}^{2}={13}^{2}$

${\left(x+2\right)}^{2}+{\left(y+4\right)}^{2}=169$

So, the equation of the circle is ${\left(x+2\right)}^{2}+{\left(y+4\right)}^{2}=169$.

## 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$\overline{)\text{Y=}}$ button.

Step 3: Enter the expression ${\left(x+2\right)}^{2}+{\left(y+4\right)}^{2}=169$ .

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 (-2, -4) and the radius is 13.