The equation of the circles which touch both the axes and the line 4x+3y=12 and have centres in the first quadrant, are

Options

A. `x^(2)+y^(2)-x-y+1=0`
B. `x^(2)+y^(2)-2x-2y+1=0`
C. `x^(2)+y^(2)-12x-12y+36=0`
D. `x^(2)+y^(2)-6x-6y+36=0`

Answer

To find the equations of the circles that touch both axes and the line \(4x + 3y = 12\) with centers in the first quadrant, we can follow these steps:

### Step 1: Understand the Circle's Properties
A circle that touches both axes will have its center at \((r, r)\), where \(r\) is the radius of the circle. This is because the distance from the center to each axis must equal the radius.

### Step 2: Find the Perpendicular Distance from the Center to the Line
The line given is \(4x + 3y - 12 = 0\). The formula for the perpendicular distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is:

\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]

Substituting \(A = 4\), \(B = 3\), \(C = -12\), and the center \((r, r)\):

\[ d = \frac{|4r + 3r - 12|}{\sqrt{4^2 + 3^2}} = \frac{|7r - 12|}{5} \]

### Step 3: Set the Distance Equal to the Radius
Since the circle touches the line, this distance \(d\) must equal the radius \(r\):

\[ \frac{|7r - 12|}{5} = r \]

### Step 4: Solve the Equation
We can split this into two cases based on the absolute value.

**Case 1:** \(7r - 12 = 5r\)

\[ 7r - 5r = 12 \implies 2r = 12 \implies r = 6 \]

**Case 2:** \(7r - 12 = -5r\)

\[ 7r + 5r = 12 \implies 12r = 12 \implies r = 1 \]

### Step 5: Find the Centers
Now we can find the centers for both values of \(r\):

1. For \(r = 6\), the center is \((6, 6)\).
2. For \(r = 1\), the center is \((1, 1)\).

### Step 6: Write the Equations of the Circles
The general equation of a circle with center \((h, k)\) and radius \(r\) is:

\[ (x - h)^2 + (y - k)^2 = r^2 \]

**For \(r = 6\):**
\[ (x - 6)^2 + (y - 6)^2 = 6^2 \implies (x - 6)^2 + (y - 6)^2 = 36 \]

Expanding this:
\[ x^2 - 12x + 36 + y^2 - 12y + 36 = 36 \implies x^2 + y^2 - 12x - 12y + 36 = 0 \]

**For \(r = 1\):**
\[ (x - 1)^2 + (y - 1)^2 = 1^2 \implies (x - 1)^2 + (y - 1)^2 = 1 \]

Expanding this:
\[ x^2 - 2x + 1 + y^2 - 2y + 1 = 1 \implies x^2 + y^2 - 2x - 2y + 1 = 0 \]

### Final Equations
The equations of the circles are:
1. \(x^2 + y^2 - 12x - 12y + 36 = 0\)
2. \(x^2 + y^2 - 2x - 2y + 1 = 0\)