Chapter 13.6, Problem 9PSA

(a)

To determine

To find: Whether the point lies inside or outside the circle for the given equation.

Answer to Problem 9PSA

The point lies on the circle.

Explanation of Solution

Given:

The given equation is $x^2+y^2=29$ .

The given point is $\left(2,5\right)$ .

Calculation:

Consider the given point is,

  $x^2+y^2=29$

Substitute $\left(2,5\right)$ in the above equation as,

  $\begin{array}{l}{\left(2\right)}^2+{\left(5\right)}^2=29\\ 29=29\end{array}$

As, the value of LHS is equal RHS, then the point lies on the circle.

(b)

To determine

To find: Whether the point lies inside or outside the circle for the given equation.

Answer to Problem 9PSA

The point lies inside the circle.

Explanation of Solution

Given:

The given equation is $x^2+y^2=100$ .

The given point is $\left(3,0\right)$ .

Calculation:

Consider the given point is,

  $x^2+y^2=100$

Substitute $\left(3,0\right)$ in the above equation as,

  $\begin{array}{l}{\left(3\right)}^2+{\left(0\right)}^2=100\\ 9<100\end{array}$

As, the value of LHS is less than RHS, then the point lies inside the circle.

(c)

To determine

To find: Whether the point lies inside or outside the circle for the given equation.

Answer to Problem 9PSA

The point lies outside the circle.

Explanation of Solution

Given:

The given equation is ${\left(x-2\right)}^2+{\left(y+5\right)}^2=15$ .

The given point is $\left(0,0\right)$ .

Calculation:

Consider the given point is,

  ${\left(x-2\right)}^2+{\left(y+5\right)}^2=15$

Substitute $\left(0,0\right)$ in the above equation as,

  $\begin{array}{l}{\left(0-2\right)}^2+{\left(0+5\right)}^2=15\\ 4+25=15\\ 29>15\end{array}$

As, the value of LHS is more than RHS, then the point lies outside the circle.

(d)

To determine

To find: Whether the point lies inside or outside the circle for the given equation.

Answer to Problem 9PSA

The point lies outside the circle.

Explanation of Solution

Given:

The given equation is $x^2+{\left(y+6\right)}^2=23$ .

The given point is $\left(-2,1\right)$ .

Calculation:

Consider the given point is,

  $x^2+{\left(y+6\right)}^2=23$

Substitute $\left(-2,1\right)$ in the above equation as,

  $\begin{array}{l}{\left(-2\right)}^2+{\left(1+6\right)}^2=23\\ 4+49>23\end{array}$

As, the value of LHS is more than RHS, then the point lies outside the circle.