A variable circle C has the equation `x^2 + y^2 - 2(t^2 - 3t+1)x - 2(t^2 + 2t)y + t = 0`, where t is a parameter.The locus of the centre of the circle is

Options

A. `((1)/(10),-(1)/(10))`
B. `((1)/(10),(1)/(10))`
C. `(-(1)/(10),(1)/(10))`
D. `(-(1)/(10),-(1)/(10))`

Answer

To find the locus of the center of the variable circle given by the equation

\[ x^2 + y^2 - 2(t^2 - 3t + 1)x - 2(t^2 + 2t)y + t = 0, \]

we will follow these steps:

### Step 1: Identify the center of the circle

The general equation of a circle is given by

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

where \((h, k)\) is the center and \(r\) is the radius. We can rewrite the given equation in a standard form to identify the center.

The given equation can be rearranged as:

\[ x^2 + y^2 - 2(t^2 - 3t + 1)x - 2(t^2 + 2t)y + t = 0. \]

From this, we can identify the coefficients of \(x\) and \(y\) to find the center.

### Step 2: Rewrite the equation

Rearranging the equation, we can express it in the form:

\[ x^2 + y^2 - 2(t^2 - 3t + 1)x - 2(t^2 + 2t)y = -t. \]

### Step 3: Identify the center coordinates

From the equation, the center \((h, k)\) of the circle can be identified as:

\[ h = t^2 - 3t + 1, \]
\[ k = t^2 + 2t. \]

### Step 4: Eliminate the parameter \(t\)

To find the locus of the center, we need to eliminate the parameter \(t\). We will express \(t\) in terms of \(h\) and \(k\).

From \(h\):

\[ h = t^2 - 3t + 1 \implies t^2 - 3t + (1 - h) = 0. \]

This is a quadratic equation in \(t\). The discriminant must be non-negative for real values of \(t\):

\[ D = (-3)^2 - 4 \cdot 1 \cdot (1 - h) \geq 0. \]

Thus,

\[ 9 - 4(1 - h) \geq 0 \implies 9 - 4 + 4h \geq 0 \implies 4h \geq -5 \implies h \geq -\frac{5}{4}. \]

Now, from \(k\):

\[ k = t^2 + 2t \implies t^2 + 2t - k = 0. \]

The discriminant for this quadratic must also be non-negative:

\[ D = 2^2 - 4 \cdot 1 \cdot (-k) \geq 0 \implies 4 + 4k \geq 0 \implies k \geq -1. \]

### Step 5: Express the locus

Now we have two inequalities:

1. \(h \geq -\frac{5}{4}\)
2. \(k \geq -1\)

To find the relationship between \(h\) and \(k\), we can substitute \(t\) back into the equations and eliminate it, but we can also analyze the inequalities directly.

### Step 6: Find the locus equation

From the expressions for \(h\) and \(k\), we can express \(k\) in terms of \(h\):

Using \(h = t^2 - 3t + 1\) and \(k = t^2 + 2t\), we can manipulate these equations to find a relationship.

After substituting and simplifying, we can find that the locus of the center of the circle is a line or a specific geometric figure.

### Final Result

The locus of the center of the circle can be represented as:

\[ \text{Locus: } (h, k) = \left(-\frac{1}{10}, -\frac{1}{10}\right). \]