Circle Equation Basics

The general equation of a circle is given by $x^2 + y^2 + 2gx + 2fy + c = 0$. In this form, the center of the circle is located at the coordinates $(-g, -f)$, and the radius is $\sqrt{g^2 + f^2 - c}$.

Using the Center Condition

We are given that the center of the circle must lie on the line $x + 2y + 3 = 0$. Let the center of the circle be $(h, k)$. According to the general equation, we have $h = -g$ and $k = -f$. Substituting these into the line equation:

$(h) + 2(k) + 3 = 0$

Now, let's express this in terms of $g$ and $f$. Substitute $h = -g$ and $k = -f$ back into the equation:

$(-g) + 2(-f) + 3 = 0$

$-g - 2f + 3 = 0$

Rearranging this equation gives us our first condition:

$g + 2f = 3$ (Equation 1)

Using Points on the Circle

The problem states that the circle passes through two specific points: $(-1, 1)$ and $(2, 1)$. We can substitute the coordinates of these points into the general circle equation $x^2 + y^2 + 2gx + 2fy + c = 0$ to get two more equations.

For the point (-1, 1):

$(-1)^2 + (1)^2 + 2g(-1) + 2f(1) + c = 0$

$1 + 1 - 2g + 2f + c = 0$

Simplifying this gives:

$-2g + 2f + c = -2$ (Equation 2)

For the point (2, 1):

$(2)^2 + (1)^2 + 2g(2) + 2f(1) + c = 0$

$4 + 1 + 4g + 2f + c = 0$

Simplifying this gives:

$4g + 2f + c = -5$ (Equation 3)

Solving for Circle Coefficients

We now have a system of three linear equations with three variables ($g$, $f$, and $c$):

• Equation 1: $g + 2f = 3$
• Equation 2: $-2g + 2f + c = -2$
• Equation 3: $4g + 2f + c = -5$

To find the values of $g$, $f$, and $c$, we can start by eliminating $c$. Subtract Equation 2 from Equation 3:

$(4g + 2f + c) - (-2g + 2f + c) = -5 - (-2)$

$4g + 2f + c + 2g - 2f - c = -5 + 2$

$6g = -3$

Solving for $g$, we get:

$g = -\frac{3}{6} = -\frac{1}{2}$

Now, substitute the value of $g$ back into Equation 1 to find $f$:

$g + 2f = 3$

$-\frac{1}{2} + 2f = 3$

$2f = 3 + \frac{1}{2}$

$2f = \frac{6}{2} + \frac{1}{2}$

$2f = \frac{7}{2}$

Solving for $f$, we get:

$f = \frac{7}{4}$

Finally, substitute the values of $g$ and $f$ into Equation 2 to find $c$:

$-2g + 2f + c = -2$

$-2(-\frac{1}{2}) + 2(\frac{7}{4}) + c = -2$

$1 + \frac{7}{2} + c = -2$

$c = -2 - 1 - \frac{7}{2}$

$c = -3 - \frac{7}{2}$

$c = -\frac{6}{2} - \frac{7}{2}$

$c = -\frac{13}{2}$

Constructing the Final Circle Equation

Now that we have found the values $g = -\frac{1}{2}$, $f = \frac{7}{4}$, and $c = -\frac{13}{2}$, we substitute them back into the general equation of the circle:

$x^2 + y^2 + 2gx + 2fy + c = 0$

$x^2 + y^2 + 2(-\frac{1}{2})x + 2(\frac{7}{4})y + (-\frac{13}{2}) = 0$

This simplifies to:

$x^2 + y^2 - x + \frac{7}{2}y - \frac{13}{2} = 0$

The options provided have integer coefficients. To match the format, we multiply the entire equation by 2:

$2 \times (x^2 + y^2 - x + \frac{7}{2}y - \frac{13}{2}) = 2 \times 0$

$2x^2 + 2y^2 - 2x + 7y - 13 = 0$

This is the equation of the circle that satisfies all the given conditions.