Chapter 1.4, Problem 41AYU

To determine

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

Answer to Problem 41AYU

The standard form of the equation of the circle is ${\left(x+1\right)}^2+{(y-3)}^2=1$

Explanation of Solution

Given:

Center $\left(-1,3\right)$ and tangent to the line $y=2$

Calculation:

A tangent to a circle is a line that touches the circle at a particular point $\left(x,y\right)$

The center of the given circle $\left(h,k\right)$ is $\left(-1,3\right)$

To plot this point in the $xy$ plane move $1$ unit to the left of $x-axis$ and $3$ units straight up.

The line $y=2$ is a horizontal line with the $y-intercept$ as $2$ and no $x-intercept$

Since the circle is a tangent to the line, it touches the line at the point $(-1,2)$ .

Therefore, the coordinates $\left(x,y\right)$ can be written as $(-1,2)$

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

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

  $r=\sqrt{{\left(-1+1\right)}^2+{\left(2-3\right)}^2}$

  $=\sqrt{1^2}$

  $=1$

Therefore, the radius of the circle is $1$

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

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

  ${\left(x+1\right)}^2+{(y-3)}^2=1$

Conclusion:

Hence, the standard form of the equation of the circle is ${\left(x+1\right)}^2+{(y-3)}^2=1$