Q.42 - Precalculus Enhanced with Graphing Utilities

Question

Find the standard form of the equation of each circle.

Center (4,-2) and tangent to the line x = 1

Step-by-Step Solution

Answer

The standard form of the equation of a circle  centered at(4,-2) and tangent to the line x=1 is ${(x+1)}^2+{(y-3)}^2=1$

Step 1. Given information

Given that the circle  centered at(4,-2) and tangent to the line x=1

 we have to find the standard form of the equation of a circle.

Step 2. Description of finding radius r of the circle

According to the given question circle center at (4,-2) and tangent to line x=1 means  a line drawn from center to the point (4,-2), perpendicular to x=1, will intersect at (1,-2). Thus  the distance from (1,-2)  to center (4,-2 ) is the radius and then the radius is $$\begin{gathered} r={\sqrt{{(1-4)}^2+{(-2-(-2\left)\right)}^2}}^2 \\ =\sqrt{{(-3)}^2+0} \\ =3 \end{gathered}$$

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  the equation of the given circle with a radius r is ${(x-4)}^2+{(y+2)}^2=9$