Q.36 - Precalculus Enhanced with Graphing Utilities

Question

Find the standard form of the equation of each circle.

Center (1, 0) and containing the point 

( -3, 2)

Step-by-Step Solution

Answer

The standard form of the equation of circle at the center(1,0) containing points (-3,2) is ${(x-2)}^2+y^2=20$ and generalized form of equation of the circle is $x^2+y^2-2x-19=0$

Step 1. Given information

Given that circle at the center (1,0) containing point at (-3,2).

 we have to formulate the standard form equation of  the circle .

Step 2. Find radius r of the circle

To find the standard form of the circle, first, we have to find the radius of the circle. Thus here center (1, 0) and contain the point (-3,2) are given, the radius is the distance between the point(-3,2) to center (1,0). So radius 

$$\begin{gathered} r=\sqrt{{(-3-1)}^2+{(2-0)}^2} \sin ce dis\tan ce between 2 points from \left(x_{1,}{y}_1\right) to\left(x_{2,}{y}_2\right) is \sqrt{{\left(x_{2-}{x}_1\right)}^2+{\left(y_{2-}{y}_1\right)}^2} \\ =\sqrt{{(-4)}^2+2^2} here \left(x_{1,}{y}_1\right) is (1,0) and \left(x_{2,}{y}_2\right) is (-3,2). \\ =\pm \sqrt{20} \\ Hence radius is \pm \sqrt{20.} \end{gathered}$$

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

and here the radius is $20$ .

Then  the standard form of the equation of the circle is  $$\begin{gathered} {(x-1)}^2+{(y-0)}^2={\left(\sqrt{20}\right)}^2 \\ {(x-1)}^2+y^2=20 \\ To get generalized form of equation open the paranthesis and solve ,then we get \\ {x}^2+y^2-2x-19=0 \end{gathered}$$