Understanding the Problem: Circle Touching Axes and a Line

The problem asks for the equation of a circle that meets specific geometric conditions: it lies entirely in the first quadrant, touches both the x-axis and the y-axis, and also touches the horizontal line given by the equation \(y - 2 = 0\).

Let's break down the conditions:

• Touches both axes in the first quadrant: If a circle in the first quadrant touches both the x-axis and the y-axis, its center must be equidistant from both axes. This distance is equal to the circle's radius. Therefore, if the radius is denoted by \(r\), the center of the circle must be at the point \((r, r)\).
• Touches the line \(y - 2 = 0\): This line is the horizontal line \(y = 2\). If the circle touches this line, the shortest distance from the center of the circle to this line must be equal to the circle's radius \(r\).

Finding the Center and Radius of the Circle

We know the center is \((r, r)\) and the radius is \(r\).

The equation of the line is \(y - 2 = 0\), which can be written in the general form \(Ax + By + C = 0\) as \(0x + 1y - 2 = 0\). Here, \(A=0\), \(B=1\), and \(C=-2\).

The distance from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\) is given by the formula:

$$ \text{Distance} = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} $$

Using the center \((r, r)\) as \((x_0, y_0)\) and the line \(0x + 1y - 2 = 0\), the distance from the center to the line \(y=2\) is:

$$ \text{Distance} = \frac{|0(r) + 1(r) - 2|}{\sqrt{0^2 + 1^2}} = \frac{|r - 2|}{\sqrt{1}} = |r - 2| $$

Since the circle touches the line, this distance must be equal to the radius \(r\).

$$ |r - 2| = r $$

We need to solve this absolute value equation for \(r\). There are two possible cases:

• Case 1: \(r - 2 = r\)
• This simplifies to \(-2 = 0\), which is not possible.

• Case 2: \(r - 2 = -r\)
• This simplifies to \(r + r = 2\), so \(2r = 2\).
• Dividing by 2, we get \(r = 1\).

Since the circle is in the first quadrant and touches the axes, the radius \(r\) must be a positive value. Thus, \(r = 1\) is the valid radius.

With \(r=1\), the center of the circle is \((1, 1)\).

Deriving the Equation of the Circle

The standard equation of a circle with center \((h, k)\) and radius \(r\) is:

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

We found the center is \((1, 1)\) and the radius is \(1\). Substituting these values into the standard equation:

$$ (x - 1)^2 + (y - 1)^2 = 1^2 $$

Now, let's expand and simplify this equation to get the general form:

$$ (x^2 - 2x + 1) + (y^2 - 2y + 1) = 1 $$

$$ x^2 - 2x + 1 + y^2 - 2y + 1 = 1 $$

Rearranging the terms and moving the constant term from the right side to the left:

$$ x^2 + y^2 - 2x - 2y + 1 + 1 - 1 = 0 $$

$$ x^2 + y^2 - 2x - 2y + 1 = 0 $$

This is the equation of the circle that touches both axes in the first quadrant and the line \(y - 2 = 0\).

Comparing with Options

Let's compare our derived equation \(x^2 + y^2 - 2x - 2y + 1 = 0\) with the given options:

• Option 1: \(x^2 + y^2 - 2x - 2y - 1 = 0\) (Does not match)
• Option 2: \(x^2 + y^2 + 2x + 2y + 1 = 0\) (Does not match)
• Option 3: \(x^2 + y^2 - 2x - 2y + 1 = 0\) (Matches)
• Option 4: \(x^2 + y^2 - 4x - 4y + 4 = 0\) (Does not match)

Our equation matches Option 3.

Revision Table: Circle Equation Basics

Additional Information: Circles Tangent to Axes

When a circle is tangent to both the x-axis and the y-axis, its center's coordinates are \(( \pm r, \pm r)\), where \(r\) is the radius. The sign depends on the quadrant the circle is in.

• First Quadrant: Center \((r, r)\)
• Second Quadrant: Center \((-r, r)\)
• Third Quadrant: Center \((-r, -r)\)
• Fourth Quadrant: Center \((r, -r)\)

In each case, the standard equation of the circle becomes simplified:

• 1st Q: \((x - r)^2 + (y - r)^2 = r^2\)
• 2nd Q: \((x + r)^2 + (y - r)^2 = r^2\)
• 3rd Q: \((x + r)^2 + (y + r)^2 = r^2\)
• 4th Q: \((x - r)^2 + (y + r)^2 = r^2\)

The additional condition of touching another line or point helps determine the value of \(r\).