Question

Find the standard form of the equation of each circle.

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

Short 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-by-step solution

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${(x-4)}^2+{(y+2)}^2=9$