To solve the problem of finding the value of \( a \) such that the points \( (2,0) \), \( (0,1) \), \( (4,0) \), and \( (0,a) \) are concyclic, we will follow these steps:
### Step 1: General Equation of a Circle
The general equation of a circle can be expressed as:
\[
x^2 + y^2 + 2gx + 2fy + c = 0
\]
where \( g \), \( f \), and \( c \) are constants.
### Step 2: Substitute the First Point (2,0)
Substituting the point \( (2,0) \) into the circle equation:
\[
2^2 + 0^2 + 2g(2) + 2f(0) + c = 0
\]
This simplifies to:
\[
4 + 4g + c = 0 \quad \text{(Equation 1)}
\]
### Step 3: Substitute the Second Point (0,1)
Now, substitute the point \( (0,1) \):
\[
0^2 + 1^2 + 2g(0) + 2f(1) + c = 0
\]
This simplifies to:
\[
1 + 2f + c = 0 \quad \text{(Equation 2)}
\]
### Step 4: Substitute the Third Point (4,0)
Next, substitute the point \( (4,0) \):
\[
4^2 + 0^2 + 2g(4) + 2f(0) + c = 0
\]
This simplifies to:
\[
16 + 8g + c = 0 \quad \text{(Equation 3)}
\]
### Step 5: Solve the Equations
Now we have three equations:
1. \( 4g + c + 4 = 0 \)
2. \( 2f + c + 1 = 0 \)
3. \( 8g + c + 16 = 0 \)
#### Step 5.1: Eliminate \( c \)
From Equation 1, we can express \( c \):
\[
c = -4g - 4
\]
Substituting this into Equation 2:
\[
2f - 4g - 4 + 1 = 0 \implies 2f - 4g - 3 = 0 \implies 2f = 4g + 3 \implies f = 2g + \frac{3}{2}
\]
#### Step 5.2: Substitute \( c \) into Equation 3
Now substituting \( c \) into Equation 3:
\[
8g - 4g - 4 + 16 = 0 \implies 4g + 12 = 0 \implies g = -3
\]
### Step 6: Find \( c \) and \( f \)
Substituting \( g = -3 \) back into the expression for \( c \):
\[
c = -4(-3) - 4 = 12 - 4 = 8
\]
Now substituting \( g = -3 \) into the expression for \( f \):
\[
f = 2(-3) + \frac{3}{2} = -6 + \frac{3}{2} = -\frac{12}{2} + \frac{3}{2} = -\frac{9}{2}
\]
### Step 7: Write the Circle Equation
Now we have:
- \( g = -3 \)
- \( f = -\frac{9}{2} \)
- \( c = 8 \)
Thus, the equation of the circle is:
\[
x^2 + y^2 - 6x - 9y + 8 = 0
\]
### Step 8: Substitute the Fourth Point (0,a)
Now, we substitute the point \( (0,a) \) into the circle equation:
\[
0^2 + a^2 - 6(0) - 9a + 8 = 0
\]
This simplifies to:
\[
a^2 - 9a + 8 = 0
\]
### Step 9: Factor the Quadratic Equation
Factoring the quadratic:
\[
(a - 1)(a - 8) = 0
\]
Thus, the solutions are:
\[
a = 1 \quad \text{or} \quad a = 8
\]
### Conclusion
The values of \( a \) are \( 1 \) and \( 8 \). Since the question asks for the value of \( a \) that makes the points concyclic, we can conclude:
\[
\text{The value of } a \text{ is } 8.
\]
