Q. 40 - Geometry

Question

Find the radii of the circles$x^2+y^2=2\text{ and }{\left(x-3\right)}^2+{\left(y-3\right)}^2=32$

Find the distance between the centers of the circles.

c. Explain why the circles must be internally tangent.

d. Sketch the graphs of the circles.

Step-by-Step Solution

Answer

Radius of $x^2+y^2=2$ is $\sqrt{2}$and the radius of${\left(x-3\right)}^2+{\left(y-3\right)}^2=32$$is \sqrt{32}$

1. The distance is $3\sqrt{2}$.
2. The distance between the centers of the circle is equal to difference of the radius. So, the circle must be internally tangent. 

The sketch is:

Step-1 – Given

The given equation is $x^2+y^2=2\text{ and }{\left(x-3\right)}^2+{\left(y-3\right)}^2=32$

Step-2 – To determine

We have to find the radii of the circles.

Step-3 – Calculation

From part b we see that the distance between the centers = $3\sqrt{2}$.

From part a, 

Radius of the first circle = $\sqrt{2}$.

Radius of the second circle = $\sqrt{32}$.

So,

$\begin{array}{l}{\text{c}}_1{c}_2=\left(r_2-r_1\right)\\ {\text{c}}_1{c}_2=\left(\sqrt{32}-\sqrt{2}\right)\\ {\text{c}}_1{c}_2=\left(4\sqrt{2}-\sqrt{2}\right)\\ {\text{c}}_1{c}_2=3\sqrt{2}\end{array}$

We see that the distance between the centers of the circle is equal to difference of the radius. So, the circles must be internally tangent.

d. Step-1 – Given

The given equations are $x^2+y^2=2$and ${\left(x-3\right)}^2+{\left(y-3\right)}^2=32$

Step-2 – To determine

We have to sketch the circles.

Step-3 – Calculation

We’ll 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 and $x^2+y^2=2$ $and {\left(x-3\right)}^2+{\left(y-3\right)}^2=32$.

Step 4: Press GRAPH button to graph the function. Then adjust the windows according to the graph.

The obtained graph is: