Question

2. What is the equation of the circle whose center is

on the line 2x + y - 6 = 0 and which is tangent
to both axes in the first quadrant?

A) x² + y² - 4x - 4y + 4 = 0
B) x² + y² - 4x - 4y - 4 = 0
C) x² + y² - 4x + 4y + 1 = 0
D) x² + y² - 2x - 2y + 2 = 0
E) x² + y² + 2x + 2y - 2 = 0


Please provide explanation

Answer 1

A) x² + y² - 4x - 4y + 4 = 0

Explanation:

You want the general form equation of the circle tangent to both axes with its center on the line 2x +y -6 = 0.

Tangent

If the circle is tangent to both the x- and y-axes, its center will be equidistant from the axes, so will lie on one of the lines y = x or y = -x. Since the given line has positive x- and y-intercepts, we presume we're interested in the circle with its center in the first quadrant.

Center

For y=x, the center point can be found by solving the system of equations ...

• y = x
• 2x +y -6 = 0

Substituting for y gives ...

2x +x -6 = 0

x -2 = 0

x = y = 2

Equation

The radius of the circle will be the x- or y-value of its center. The standard form equation of a circle with radius r and center (h, k) is ...

(x -h)² +(y -k)² = r²

For h = k = r = 2, this is ...

(x -2)² +(y -2)² = 2²

x² -4x +4 +y² -4y +4 -4 = 0 . . . . . expand, subtract r²

x² +y² -4x -4y +4 = 0 . . . . . . . . simplify to general form

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Additional comment

There is also a 4th quadrant circle tangent to both axes with its center on the given line. Its center is (6, -6) and its equation is ...

x² +y² -12x +12y +36 = 0 . . . . . . not an answer choice