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We are given the equation of the circle \( L \):

\[
L: 2x^2 + 2y^2 - 4x + 3py - 30 = 0
\]

The circle \( M \) is concentric with circle \( L \), meaning both circles have the same center, but circle \( M \) has a radius twice the radius of circle \( L \). Additionally, circle \( M \) passes through the point \((-2, 1)\).

### Step 1: Simplify the equation of circle \( L \)
First, we rewrite the equation of circle \( L \) in standard form by dividing through by 2:

\[
x^2 + y^2 - 2x + \frac{3p}{2}y - 15 = 0
\]

We now complete the square to find the center and radius of circle \( L \).

- For the \(x\)-terms: \(x^2 - 2x\) becomes \((x - 1)^2 - 1\).
- For the \(y\)-terms: \(y^2 + \frac{3p}{2}y\). To complete the square, we add and subtract \(\left(\frac{3p}{4}\right)^2 = \frac{9p^2}{16}\), so this becomes:

\[
\left(y + \frac{3p}{4}\right)^2 - \frac{9p^2}{16}
\]

Thus, the equation of circle \( L \) becomes:

\[
(x - 1)^2 + \left(y + \frac{3p}{4}\right)^2 = 15 + 1 + \frac{9p^2}{16}
\]

This gives the center of circle \( L \) as \( (1, -\frac{3p}{4}) \) and its radius squared as:

\[
r_L^2 = 16 + \frac{9p^2}{16}
\]

### Step 2: Find the equation of circle \( M \)
Since circle \( M \) is concentric with \( L \), its center is also \( (1, -\frac{3p}{4}) \), and its radius is twice that of \( L \). Circle \( M \) also passes through the point \((-2, 1)\).

The equation of circle \( M \) will have the form:

\[
(x - 1)^2 + \left(y + \frac{3p}{4}\right)^2 = R_M^2
\]

To find \( R_M \), we substitute the point \((-2, 1)\) into this equation:

\[
(-2 - 1)^2 + \left(1 + \frac{3p}{4}\right)^2 = R_M^2
\]

Now, we can proceed to solve for \( p \) and \( R_M \).

Let me calculate that for you.After substituting the point \((-2, 1)\) into the equation of circle \( M \), we obtain the following expression for \( R_M^2 \):

\[
R_M^2 = \frac{(3p + 4)^2}{16} + 9
\]

This gives us the radius squared of circle \( M \) in terms of \( p \). To find the equation of the circle \( M \), we now need to finalize the value of \( p \), if additional conditions or steps are given.

Do you want more details on how to proceed further with this value, or should I assist you with more specific steps?

Here are 5 related questions for further exploration:
1. How can we further simplify the equation of circle \( M \) given \( p \)?
2. What is the geometric interpretation of two concentric circles?
3. How do we solve for \( p \) if more information is available?
4. Can the method for completing the square be applied to other conic sections?
5. What are the applications of concentric circles in real-life geometry problems?

**Tip:** When solving geometric problems involving circles, always try to express the equation in its simplest form to identify the center and radius easily.

Diketahui lingkaran L: 2x^2 + 2y^2 - 4x + 3py - 30 = 0 melalui titik (-2,1). Lingkaran M sepusat dengan lingkaran L, tetapi jari-jarinya dua kali lingkaran L. Tentukan persamaan lingkaran M.

Solve the equation for circle M, which is concentric with circle L and has a radius twice as large. Learn how to use completing the square to find the center and radius of a circle and apply geometric theorems in solving this problem. Suitable for students in Grades 10-12.

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