Q.39 - Precalculus Enhanced with Graphing Utilities

Question

Find the standard form of the equation of each circle.

Circle with endpoints of a diameter at (1, 4) and (-3, 2)

Step-by-Step Solution

Answer

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

Step 1. Given information

  Here given that circle with endpoints of a diameter at (1, 4) and (-3, 2).

 we have to find the standard for the equation of  the given circle,

Step 2. Explanation about finding radius r of the circle

Here circle endpoint with a diameter 

(1, 4) and (-3, 2) is given.  The center of the circle is the midpoints  of the circle  endpoints (1,4) and (-3,2)  , hence the center is $(\frac{1-3}{2},\frac{2+4}{2})=(-1,3), \sin ce midpoint(x, y) is between (x_1,y_1) and (x_2,y_2) and then (x,y)= (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

 We know that radius is the distance between the endpoint to the center  , thus radius  r is $$\begin{gathered} r=\sqrt{{(1-(-1)}^2+{(4-3)}^2} \\ =\sqrt{{\left(2\right)}^2+1} \\ =\sqrt{4+1} \\ =\sqrt{5} \end{gathered}$$

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

Hence  equation of the given circle is $$\begin{gathered} {(x-(-1\left)\right)}^2+{(y-3)}^2={\sqrt{5}}^2 \\ {(x+1)}^2+{(y-3)}^2=5 \end{gathered}$$