Chapter 1.4, Problem 37AYU

To determine

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

Answer to Problem 37AYU

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

Explanation of Solution

Given:

Center: $\left(2,3\right)$

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 $x-axis$ , it intercepts the $x-axis$ at some point $\left(x,y\right)$ .

The center of the circle $\left(h,k\right)$ is given as $\left(2,3\right)$ .

In order to get to the point $\left(2,3\right)$ in the xy-plane, go $2$ units to the right of $x-axis$ and thenstraight up $3$ units.

So, the x-coordinate of the center of the circle will be the $x$ intercept of the circle.

Therefore, the coordinates $\left(x,y\right)$ can be written as $\left(2,0\right)$ .

If $\left(x,y\right)$ represent 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 the points $\left(x,y\right)$ and $\left(h,k\right)$ .

  $\begin{array}{l}\sqrt{{\left(x-h\right)}^2+{\left(y-k\right)}^2}=r\\ r=\sqrt{{\left(2-2\right)}^2+{\left(3-0\right)}^2}\\ =\sqrt{3^2}\\ =3\end{array}$

Therefore, the radius of the circle is $3$ .

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

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

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

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