We are tasked with finding the minimum distance between two circles, \( C_1 \) and \( C_2 \), given their centers and radii. Let us proceed step by step:
1. Given Information:
The centers and radii of the circles are:
\( C_1(8, 2), \quad r_1 = 1 \)
\( C_2(2, 6), \quad r_2 = 2 \)
2. Distance Between the Centers:
The distance between the centers \( C_1 \) and \( C_2 \) is given by the Euclidean distance formula:
\( C_1C_2 = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
3. Minimum Distance Between the Circles:
The minimum distance between the two circles is the distance between their centers minus the sum of their radii:
\( |z_1 - z_2| = C_1C_2 - (r_1 + r_2) \)
Substitute \( C_1C_2 = 10 \), \( r_1 = 1 \), and \( r_2 = 2 \):
\( |z_1 - z_2| = 10 - (1 + 2) \)
\( |z_1 - z_2| = 10 - 3 = 7 \)
Final Answer:
The minimum distance between the two circles is \( \boxed{7} \).
Let $C$ be the circle $x^2 + (y - 1)^2 = 2$, $E_1$ and $E_2$ be two ellipses whose centres lie at the origin and major axes lie on the $x$-axis and $y$-axis respectively. Let the straight line $x + y = 3$ touch the curves $C$, $E_1$, and $E_2$ at $P(x_1, y_1)$, $Q(x_2, y_2)$, and $R(x_3, y_3)$ respectively. Given that $P$ is the mid-point of the line segment $QR$ and $PQ = \frac{2\sqrt{2}}{3}$, the value of $9(x_1 y_1 + x_2 y_2 + x_3 y_3)$ is equal to
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: