Find the equation of the circle which touch the line 2x-y=1 at (1,1) and line 2x+y=4

Options

A. `x+3y=2`
B. `x+2y=3`
C. `x+y=2`
D. none of these

Explanation

To find the equation of the circle that touches the line \(2x - y = 1\) at the point \((1, 1)\) and also touches the line \(2x + y = 4\), we can follow these steps:

### Step 1: Identify the center of the circle
Let the center of the circle be \(C(h, k)\). Since the circle touches the line \(2x - y = 1\) at the point \((1, 1)\), the distance from the center \(C(h, k)\) to the line must equal the radius \(r\) of the circle.

### Step 2: Calculate the distance from the center to the line
The distance \(d\) from a point \((x_1, y_1)\) to a line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \(2x - y - 1 = 0\) (where \(A = 2\), \(B = -1\), and \(C = -1\)), the distance from the center \(C(h, k)\) to the line is:
\[ d = \frac{|2h - k - 1|}{\sqrt{2^2 + (-1)^2}} = \frac{|2h - k - 1|}{\sqrt{5}} \]

### Step 3: Calculate the distance from the center to the point of tangency
The distance from the center \(C(h, k)\) to the point of tangency \((1, 1)\) is:
\[ \sqrt{(h - 1)^2 + (k - 1)^2} \]

### Step 4: Set the distances equal
Since the distance from the center to the line equals the distance from the center to the point of tangency (which is the radius \(r\)), we have:
\[ \frac{|2h - k - 1|}{\sqrt{5}} = \sqrt{(h - 1)^2 + (k - 1)^2} \]

### Step 5: Square both sides
Squaring both sides gives:
\[ \frac{(2h - k - 1)^2}{5} = (h - 1)^2 + (k - 1)^2 \]
Multiplying through by 5:
\[ (2h - k - 1)^2 = 5((h - 1)^2 + (k - 1)^2) \]

### Step 6: Expand both sides
Expanding both sides:
\[ (2h - k - 1)^2 = 4h^2 - 4hk + k^2 - 4h + 2k + 1 \]
And for the right side:
\[ 5((h - 1)^2 + (k - 1)^2) = 5(h^2 - 2h + 1 + k^2 - 2k + 1) = 5h^2 + 5k^2 - 10h - 10k + 10 \]

### Step 7: Set the expanded equations equal
Setting the two expansions equal gives:
\[ 4h^2 - 4hk + k^2 - 4h + 2k + 1 = 5h^2 + 5k^2 - 10h - 10k + 10 \]

### Step 8: Rearrange the equation
Rearranging terms leads to:
\[ 0 = h^2 + 4hk + 4h + 4k - 9 \]

### Step 9: Substitute \(h\) and \(k\) with \(x\) and \(y\)
This equation represents the locus of the center of the circle. We can replace \(h\) and \(k\) with \(x\) and \(y\):
\[ x^2 + 4xy + 4x + 4y - 9 = 0 \]

### Step 10: Final equation of the circle
The final equation of the circle is:
\[ x^2 + 4xy + 4x + 4y - 9 = 0 \]