Chapter 1.4, Problem 38AYU

To determine

To find: the standard form of the equation of the circle

Answer to Problem 38AYU

The standard form of the equation of the circle is ${\left(x+3\right)}^2+{\left(y-k\right)}^2=9$

Explanation of Solution

Given:

Center: $(-3,1)$

Formula Used:

The equation of the circle is $\sqrt{{\left(x-h\right)}^2+{\left(y-k\right)}^2}=r$

Calculation:

A tangent to a circle is a line that touches the circle at a particular point. Since the given circle is a tangent to the $y-axis$ . It intercepts y-axis at some point $\left(x,y\right)$ .

The center of the circle $\left(h,k\right)$ is given as $\left(-3,1\right)$ . Thus, the y-coordinate of the center of the circle will be the y-intercept of the circle. The coordinates $\left(x,y\right)$ can be written as $\left(0,1\right)$ .

If $\left(x,y\right)$ represents the coordinates of any point on the circle with radius $r$ and center $\left(h,k\right)$ then by the distance formula, the radius of the circle is equivalent to the distance between thepoints $\left(x,y\right)$ and $\left(h,k\right)$ .

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

Substitute $x=0,y=1,h=-3,k=1$

  $\begin{array}{l}r=\sqrt{{\left(0-\left(-3\right)\right)}^2+{\left(1-1\right)}^2}\\ =\sqrt{9+0}\\ =3\end{array}$

The radius of the circle is $3$ .

The standard form of the circle is given by ${\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2$ .

Substitute for $h,\text{ k},andr$ .

  $\begin{array}{l}{\left[x-\left(-3\right)\right]}^2+{\left(y-1\right)}^2={\left(3\right)}^2\\ {\left(x+3\right)}^2+{\left(y-k\right)}^2=9\end{array}$

Therefore, the standard form of the equation of the circle is ${\left(x+3\right)}^2+{\left(y-k\right)}^2=9$ .

Conclusion:The standard form of the equation of the circle is ${\left(x+3\right)}^2+{\left(y-k\right)}^2=9$