The centre of a circle is (x - 2, x+1) and it passes through the points (4,4). Find the value (or values) of x, if the diameter of the circle is of length `2sqrt(5)` units.

Options

A. 1 or 3
B. -1 or 4
C. 5 or 4
D. 3 or -2

Answer

To solve the problem step by step, we need to find the value(s) of \( x \) for the given conditions.

### Step 1: Identify the center and radius of the circle
The center of the circle is given as \( (x - 2, x + 1) \) and the diameter is given as \( 2\sqrt{5} \).

To find the radius \( r \):
\[ r = \frac{\text{Diameter}}{2} = \frac{2\sqrt{5}}{2} = \sqrt{5} \]

### Step 2: Use the distance formula
The circle passes through the point \( (4, 4) \). The distance from the center to this point must equal the radius.

Using the distance formula:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
where \( (x_1, y_1) \) is the center \( (x - 2, x + 1) \) and \( (x_2, y_2) \) is the point \( (4, 4) \).

Substituting the values:
\[ \sqrt{(4 - (x - 2))^2 + (4 - (x + 1))^2} = \sqrt{5} \]

### Step 3: Simplify the equation
First, simplify the expressions inside the square root:
\[ 4 - (x - 2) = 6 - x \quad \text{and} \quad 4 - (x + 1) = 3 - x \]
Thus, the equation becomes:
\[ \sqrt{(6 - x)^2 + (3 - x)^2} = \sqrt{5} \]

### Step 4: Square both sides
To eliminate the square root, square both sides:
\[ (6 - x)^2 + (3 - x)^2 = 5 \]

### Step 5: Expand the squares
Expanding both squares:
\[ (6 - x)^2 = 36 - 12x + x^2 \]
\[ (3 - x)^2 = 9 - 6x + x^2 \]
Combining these gives:
\[ 36 - 12x + x^2 + 9 - 6x + x^2 = 5 \]
\[ 2x^2 - 18x + 45 = 5 \]

### Step 6: Rearrange the equation
Rearranging gives:
\[ 2x^2 - 18x + 40 = 0 \]

### Step 7: Simplify the quadratic equation
Dividing the entire equation by 2:
\[ x^2 - 9x + 20 = 0 \]

### Step 8: Factor the quadratic equation
Factoring:
\[ (x - 4)(x - 5) = 0 \]
Thus, the solutions for \( x \) are:
\[ x = 4 \quad \text{or} \quad x = 5 \]

### Final Answer
The values of \( x \) are \( 4 \) and \( 5 \).

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