Chapter 13.6, Problem 8PSA

(a)

To determine

To find: The equation of the circle for the given condition.

Answer to Problem 8PSA

The equation of the circle is $x^2+y^2=25$ .

Explanation of Solution

Given:

The given condition is that the centre is at the origin and the circle passes through the points $\left(0,-5\right)$ .

Calculation:

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.

Consider the centre is at the origin that is $\left(h,k\right)=0$ .

Consider the formula for the distance between the two points is,

  $r=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The centre of the circle is at $\left(0,0\right)$ and the radius of the circle by the distance formula is obtained as,

  $\begin{array}{c}r=\sqrt{{\left(-5-0\right)}^2+{\left(0-0\right)}^2}\\ =5\end{array}$

Then, substitute the values to obtain the equation for the circle as,

  $\begin{array}{l}{\left(x-0\right)}^2+{\left(y-0\right)}^2=5^2\\ x^2+y^2=25\end{array}$

(b)

To determine

To find: The equation of the circle for the given condition.

Answer to Problem 8PSA

The equation of the circle is ${\left(x-3\right)}^2+{\left(y-13\right)}^2=169$ .

Explanation of Solution

Given:

The given condition is that the end points of the diameter is at $\left(-2,1\right)$ and $\left(8,25\right)$ .

Calculation:

Consider the mid-point of the two point is obtained as,

  $\begin{array}{c}M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ =\left(\frac{-2+8}{2},\frac{1+25}{2}\right)\\ =\left(3,13\right)\end{array}$

The centre is at is $\left(h,k\right)=\left(3,13\right)$ .

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.

Consider the formula for the distance between the two points is,

  $r=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The centre of the circle is at $\left(3,13\right)$ and the radius of the circle by the distance formula is obtained as,

  $\begin{array}{c}r=\sqrt{{\left(-2-3\right)}^2+{\left(1-13\right)}^2}\\ =\sqrt{25+144}\\ =\sqrt{169}\\ =13\end{array}$

Then, substitute the values to obtain the equation for the circle as,

  $\begin{array}{l}{\left(x-3\right)}^2+{\left(y-13\right)}^2={13}^2\\ {\left(x-3\right)}^2+{\left(y-13\right)}^2=169\end{array}$

(c)

To determine

To find: The equation of the circle for the given condition.

Answer to Problem 8PSA

The equation of the circle is ${\left(x+1\right)}^2+{\left(y-7\right)}^2=50$ .

Explanation of Solution

Given:

The given condition is that the centre is at $\left(-1,7\right)$ and the circle passes through the origin.

Calculation:

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.

Consider the centre isat $\left(h,k\right)=\left(-1,7\right)$ .

Consider the formula for the distance between the two points is,

  $r=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The centre of the circle is at $\left(-1,7\right)$ and the radius of the circle by the distance formula is obtained as,

  $\begin{array}{c}r=\sqrt{{\left(-1-0\right)}^2+{\left(7-0\right)}^2}\\ =\sqrt{50}\end{array}$

Then, substitute the values to obtain the equation for the circle as,

  $\begin{array}{l}{\left(x-\left(-1\right)\right)}^2+{\left(y-7\right)}^2={\sqrt{50}}^2\\ {\left(x+1\right)}^2+{\left(y-7\right)}^2=50\end{array}$

(d)

To determine

To find: The equation of the circle for the given condition.

Answer to Problem 8PSA

The equation of the circle is ${\left(x-2\right)}^2+{\left(y+3\right)}^2=10$ .

Explanation of Solution

Given:

The given condition is that the centre is at $\left(2,-3\right)$ and the circle passes through the point $\left(3,10\right)$ .

Calculation:

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.

Consider the centre is at $\left(h,k\right)=\left(2,-3\right)$ .

Consider the formula for the distance between the two points is,

  $r=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The centre of the circle is at $\left(2,-3\right)$ and the radius of the circle by the distance formula is obtained as,

  $\begin{array}{c}r=\sqrt{{\left(2-3\right)}^2+{\left(-3-0\right)}^2}\\ =\sqrt{10}\end{array}$

Then, substitute the values to obtain the equation for the circle as,

  $\begin{array}{l}{\left(x-2\right)}^2+{\left(y-\left(-3\right)\right)}^2={\sqrt{10}}^2\\ {\left(x-2\right)}^2+{\left(y+3\right)}^2=10\end{array}$