Chapter 13.6, Problem 6PSA

(a)

To determine

To find: The find the type of circle that is represented by the given equation.

Answer to Problem 6PSA

The circle is the point circle.

Explanation of Solution

Given:

The given equation is ${\left(x-3\right)}^2+{\left(y+1\right)}^2=0$ .

Calculation:

Consider the given equation is,

  ${\left(x-3\right)}^2+{\left(y+1\right)}^2=0$

Consider the formula for the equation of the circle is of the form as,

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

Here, $\left(h,k\right)$ is the centre and $r$ is the radius.

As the radius of the circle is 0 it represent the point circle.

(b)

To determine

To find: The find the type of circle that is represented by the given equation.

Answer to Problem 6PSA

The circle is an imaginary circle.

Explanation of Solution

Given:

The given equation is ${\left(x+5\right)}^2+y^2=-100$ .

Calculation:

Consider the given equation is,

  ${\left(x+5\right)}^2+y^2=-100$

Consider the formula for the equation of the circle is of the form as,

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

Here, $\left(h,k\right)$ is the centre and $r$ is the radius.

As the radius of the circle is $\sqrt{-100}=10i$ thus, it represent animaginary circle.