Q.Find the equation of a circle concentric with the circle $x^2 + y^2 - 6x + 12y + 15 = 0$ and has double of its area.

The key idea is to use the standard form of a circle's equation to find its center and radius, then apply the conditions for concentricity and area to determine the new circle's parameters.

1. Find the center and radius of the given circle:

   The equation $x^2 + y^2 - 6x + 12y + 15 = 0$ is in the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.

   Comparing coefficients, $2g = -6 \implies g = -3$ and $2f = 12 \implies f = 6$.

   The center $(h, k)$ is $(-g, -f) = (3, -6)$.

   The radius $r_1 = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + (6)^2 - 15} = \sqrt{9 + 36 - 15} = \sqrt{30}$.

2. Determine the radius of the new circle:

   The new circle is concentric, so its center is also $(3, -6)$.

   Its area is double that of the given circle. Let the new radius be $r_2$.

   Area of given circle $A_1 = \pi r_1^2 = \pi (\sqrt{30})^2 = 30\pi$.

   Area of new circle $A_2 = 2A_1 = 2(30\pi) = 60\pi$. …

To find the equation of a circle, we fundamentally need two pieces of information: its center coordinates $(h, k)$ and its radius $r$. The standard form of a circle's equation directly provides these:

The problem gives us an initial circle in its general form, $x^2 + y^2 - 6x + 12y + 15 = 0$. Our first step is to convert this into the standard form to extract its center and radius.

The new circle we need to find is "concentric" with the given circle. This means they share the exact same center. The second condition, "has double of its area," will help us determine the radius of this new circle.

Let's proceed step-by-step:

1. Find the center and radius of the given circle.

   The given equation is $x^2 + y^2 - 6x + 12y + 15 = 0$.

   To convert this to the standard form $(x-h)^2 + (y-k)^2 = r^2$, we use the method of completing the square for the $x$ terms and $y$ terms separately.

   Group the $x$ terms and $y$ terms:

   $(x^2 - 6x) + (y^2 + 12y) + 15 = 0$

   To complete the square for $x^2 - 6x$, we add $(\frac{-6}{2})^2 = (-3)^2 = 9$.

   To complete the square for $y^2 + 12y$, we add $(\frac{12}{2})^2 = (6)^2 = 36$.

   We add and subtract these values to maintain the equality:

   $(x^2 - 6x + 9 - 9) + (y^2 + 12y + 36 - 36) + 15 = 0$

   Now, rewrite the perfect square trinomials:

   $(x-3)^2 - 9 + (y+6)^2 - 36 + 15 = 0$

   Combine the constant terms:

   $(x-3)^2 + (y+6)^2 - 9 - 36 + 15 = 0$

   $(x-3)^2 + (y+6)^2 - 45 + 15 = 0$

   $(x-3)^2 + (y+6)^2 - 30 = 0$

   Move the constant term to the right side:

   $(x-3)^2 + (y+6)^2 = 30$

   Comparing this to the standard form $(x-h)^2 + (y-k)^2 = r^2$:

   The center of the given circle is $(h, k) = (3, -6)$.

   The square of its radius is $r_1^2 = 30$.

   Tip

   For a general equation of a circle $x^2 + y^2 + 2gx + 2fy + c = 0$, the center is $(-g, -f)$ and the radius squared is $r^2 = g^2 + f^2 - c$.

   In our case, $2g = -6 \implies g = -3$, and $2f = 12 \implies f = 6$, and $c = 15$.

   Center: $(-(-3), -(6)) = (3, -6)$.

   Radius squared: $(-3)^2 + (6)^2 - 15 = 9 + 36 - 15 = 45 - 15 = 30$. …