Question

Find the standard form of the equation of each circle.

Center (-1, 3) and tangent to the line y = 2

Short answer

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

Step-by-step solution

Step 1. Given information

Given that the circle centered at(-1,3) and tangent to the line y=2

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

Step 2. Description of finding  radius r of the circle

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

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

Here radius r=1 then the equation of the given circle becomes$$\begin{gathered} {(x-(-1\left)\right)}^2+{(y-3)}^2=1^2 \\ {(x+1)}^2+{(y-3)}^2=1 \end{gathered}$$