Chapter 1.4, Problem 40AYU

To determine

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

Answer to Problem 40AYU

The standard form of the equation of the circle is ${\left(x-2\right)}^2+{(y-2)}^2=5$

Explanation of Solution

Given:

Midpoint: $\left(4,3\right)$ and $\left(0,1\right)$

Calculation:

The center of the circle $\left(h,k\right)$ is equidistant from any point on the circle

The center of the given circle is the midpoint of the line joining the points $\left(4,3\right)$ and $\left(0,1\right)$

Use the midpoint formula to find the center of the circle

  $M=\left(x,y\right)=\left(\frac{x_{1+}{x}_2}{2},\frac{y_1+y_2}{2}\right)$

Substitute the values $4$ for $x_1$ , $3$ for $y_1$ , $0$ for $x_2$ and $1$ for $y_2$

  $\left(x,y\right)=\left(\frac{4+0}{2},\frac{3+1}{2}\right)$

Evaluate

  $\left(\frac{4+0}{2},\frac{3+1}{2}\right)=\left(\frac{4}{2},\frac{4}{2}\right)$

  $=\left(2,2\right)$

Thus, the center of the circle is $\left(2,2\right)$

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

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

Consider the point $\left(4,3\right)$

Substitute $x=4,h=2,y=3,k=2$

  $r=\sqrt{{\left(4-2\right)}^2+{\left(3-2\right)}^2}$

  $=\sqrt{4+1}$

  $=\sqrt{5}$

The radius is $\sqrt{5}$

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

Substitute the values of $h,k$ and $r$

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

  ${\left(x-2\right)}^2+{(y-2)}^2=5$

Conclusion:

Hence, the standard form of the equation of the circle is ${\left(x-2\right)}^2+{(y-2)}^2=5$