Q.40 - Precalculus Enhanced with Graphing Utilities

Question

Find the standard form of the equation of each circle.

Circle with endpoints of a diameter at (4,3) and (0,1)

Step-by-Step Solution

Answer

The standard form of the equation of a circle with endpoints of a diameter at (4,3) and (0, 1) is ${(x-2)}^2+{(y-2)}^2=5$

Step 1. Given information

  Here given that circle with endpoints of a diameter at (4,3) and (0,1).

We need to find the standard form of the equation of a circle.

Step 2. Explanation about finding radius r of the circle

According to the question circle endpoint with a diameter is

(4,3) and (0,1) .  The center of the circle is the midpoints  of the circle  endpoints (4,3) and (0,1) , hence the center is $(\frac{4+0}{2},\frac{3+1}{2})=(2,2), we know that midpoint(x, y) is between (x_1,y_1) and (x_2,y_2) and then (x,y)= (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$ The radius is the distance between the endpoint to the center, thus radius r is 

 $$\begin{gathered} r=\sqrt{{(4-2)}^2+{(3-2)}^2} \\ =\sqrt{{\left(2\right)}^2+1} \\ =\sqrt{4+1} \\ =\sqrt{5} \end{gathered}$$

Step 3. Formation of standard of the equation of the circle

The standard form of the equation of a circle is of radius r with center at the

origin (h, k)  is ${(x-h)}^2+{(y-k)}^2=r^2$

Thus  the equation of the given circle is $$\begin{gathered} {(x-2)}^2+{(y-2)}^2={\sqrt{5}}^2 \\ {(x-2)}^2+{(y-2)}^2=5 \end{gathered}$$