Chapter 1.4, Problem 39AYU

To determine

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

Answer to Problem 39AYU

The standard form of the equation of the circle is ${\left(x+1\right)}^2+{(y-3)}^2=5$

Explanation of Solution

Given:

The diameter as $\left(1,4\right)$ and $\left(-3,2\right)$

Calculation:

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

The endpoints on the diameter of the circle are given as $\left(1,4\right)$ and $\left(-3,2\right)$ . So, the center is the midpoint of the line joining the points $\left(1,4\right)$ and $\left(-3,2\right)$ .

The coordinates of the center can be obtained using the midpoint formula.

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

  $\left(h,k\right)=\left(\frac{1+-3}{2},\frac{1+-3}{2}\right)$

  $h=\frac{1+-3}{2}=-1$

  $k=\frac{1+-3}{2}=3$

The center is $(-1,3)$ .

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 $\left(1,4\right)$ on the circle with the center $\left(-1,3\right)$

  $r=\sqrt{{\left(1+1\right)}^2+{\left(4-3\right)}^2}$

  $=\sqrt{2^2+1^2}$

  $=\sqrt{5}$

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 $r$

  ${\left(x+1\right)}^2+{(y-3)}^2={\sqrt{5}}^2$

  ${\left(x+1\right)}^2+{(y-3)}^2=5$

Conclusion:

Hence, the standard form of the equation of the circle is ${\left(x+1\right)}^2+{(y-3)}^2=5$