From 68840e5464bb10d2e6c8bc49f1221853ca4a0511 Mon Sep 17 00:00:00 2001
From: SinnoLn <ok373737@naver.com>
Date: Sat, 3 Oct 2026 18:34:14 +0900
Subject: [PATCH] 1401. Circle and Rectangle Overlapping

---
 ...01. Circle and Rectangle Overlapping.java" | 24 +++++++++++++++++++
 1 file changed, 24 insertions(+)
 create mode 100644 "leetcode3/\354\235\264\354\247\204\355\235\254/1401. Circle and Rectangle Overlapping.java"

diff --git "a/leetcode3/\354\235\264\354\247\204\355\235\254/1401. Circle and Rectangle Overlapping.java" "b/leetcode3/\354\235\264\354\247\204\355\235\254/1401. Circle and Rectangle Overlapping.java"
new file mode 100644
index 00000000..75d6eab8
--- /dev/null
+++ "b/leetcode3/\354\235\264\354\247\204\355\235\254/1401. Circle and Rectangle Overlapping.java"	
@@ -0,0 +1,24 @@
+/* 
+
+1. 아이디어 : 기하학의 원리 활용
+            직사각형은 x,y축과 평행하므로 원과 가장 가까운 점을 찾는다.
+            그 점과 원의 중심사이의 거리가 반지름보다 크다면 겹치지않고, 이하라면 겹침
+
+2. 시간복잡도 : O(1)
+
+3. 자료구조/알고리즘 : 기하학
+
+ */
+
+class Solution {
+    public boolean checkOverlap(int radius, int xCenter, int yCenter, int x1, int y1, int x2, int y2) {
+        // 겹치면 true, 아니면 false
+        
+        int y = Math.max(y1, Math.min(y2, yCenter));
+        int x = Math.max(x1, Math.min(x2, xCenter));
+
+        if(((y-yCenter)*(y-yCenter) + (x-xCenter)*(x-xCenter))<=radius*radius) return true;
+
+        return false;
+    }
+}
\ No newline at end of file