You will still suffer TLE ,for that mantain your heap of size K at max
The following problem can be easily solved by using heaps.
For every query of type 1, insert elements until the size of the heap becomes ‘k’.
Then for every query of type 1 after reaching the size k for heap(max-heap) we will check if the current element is smaller than the root of the heap or not. If it is not smaller then we ignore it else we remove the root of the heap and push the new element in the heap. (What this will do is maintain a heap of size k which will contain k nearest coordinates for the dean) .
For every query of type 2 just print the root of the heap.