Equation of a circle with centre `C (h,k) and ` radius 5 such that `h ^(2)- 3h + 2 =0 and k ^(2) + 5k - 6 =0` is

Options

A. `x ^(2) + y ^(2) - 2x - 4y =0`
B. `x ^(2) + y ^(2) - 2x -2y -6=0`
C. `x ^(2) +y ^(2) - 4x - 2y - 20=0`
D. `x ^(2) + y ^(2) - 4x - 6y -12=0.`

Answer

To find the equation of a circle with center \( C(h, k) \) and radius 5, given the conditions \( h^2 - 3h + 2 = 0 \) and \( k^2 + 5k - 6 = 0 \), we will follow these steps:

### Step 1: Solve for \( h \)
We start with the equation:
\[ h^2 - 3h + 2 = 0 \]
This is a quadratic equation that can be factored. We need to find two numbers that multiply to \( 2 \) (the constant term) and add up to \( -3 \) (the coefficient of \( h \)).

Factoring gives us:
\[ (h - 1)(h - 2) = 0 \]
Setting each factor to zero gives:
\[ h - 1 = 0 \quad \Rightarrow \quad h = 1 \]
\[ h - 2 = 0 \quad \Rightarrow \quad h = 2 \]
Thus, the possible values for \( h \) are \( 1 \) and \( 2 \).

### Step 2: Solve for \( k \)
Next, we solve the equation:
\[ k^2 + 5k - 6 = 0 \]
Again, we will factor this quadratic. We need two numbers that multiply to \( -6 \) and add up to \( 5 \).

Factoring gives us:
\[ (k + 6)(k - 1) = 0 \]
Setting each factor to zero gives:
\[ k + 6 = 0 \quad \Rightarrow \quad k = -6 \]
\[ k - 1 = 0 \quad \Rightarrow \quad k = 1 \]
Thus, the possible values for \( k \) are \( -6 \) and \( 1 \).

### Step 3: Determine the combinations of \( (h, k) \)
Now we can form the combinations of \( (h, k) \):
1. \( (1, 1) \)
2. \( (1, -6) \)
3. \( (2, 1) \)
4. \( (2, -6) \)

### Step 4: Write the equation of the circle
The standard form of the equation of a circle is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \( r \) is the radius. Given that the radius is \( 5 \), we have \( r^2 = 25 \).

We will write the equations for each combination:

1. For \( (1, 1) \):
\[ (x - 1)^2 + (y - 1)^2 = 25 \]

2. For \( (1, -6) \):
\[ (x - 1)^2 + (y + 6)^2 = 25 \]

3. For \( (2, 1) \):
\[ (x - 2)^2 + (y - 1)^2 = 25 \]

4. For \( (2, -6) \):
\[ (x - 2)^2 + (y + 6)^2 = 25 \]

### Step 5: Expand one of the equations
Let's expand the equation for \( (2, 1) \):
\[ (x - 2)^2 + (y - 1)^2 = 25 \]
Expanding gives:
\[ (x^2 - 4x + 4) + (y^2 - 2y + 1) = 25 \]
Combining like terms:
\[ x^2 + y^2 - 4x - 2y + 5 = 25 \]
Subtracting 25 from both sides:
\[ x^2 + y^2 - 4x - 2y - 20 = 0 \]

### Final Result
The equation of the circle is:
\[ x^2 + y^2 - 4x - 2y - 20 = 0 \]