$A(0,~1)$ and $P$ that are points on a Cartesian coordinates system, satisfying these conditions.
A) Point P is on Quadrant 1.
B) For some point $Q$ on the $x$ -axis, $\overline{AQ}+\overline{PQ}\le6.$
Let $D$ be the area shown by all possible locations of $P.$
A point $(x,~y)$ belongs to $D.$ The maximum of $x+y$ is $p+q\sqrt{2}$ where $p,~q$ are integers.
Find the value of $p+q.$
This problem is a part of <Grade 11 CSAT Mock test> series .
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Let $A'(0,~-1).$
Then it's clear that
$\overline{AQ}+\overline{PQ}=\overline{A'Q}+\overline{PQ}\ge\overline{A'P}.$
And since there exists a point $Q$ such that $\overline{AQ}+\overline{PQ}\le6,$ we can say that
$\overline{A'P}\le6.$
Then point P should be in a circle that has a radius of $6$ and a center of $A'.$
Let $x+y=k$ and you see that $y=-x+k.$
Then $k$ is the $y$ -intercept of $y=-x+k.$
Move the linear graph up and down and you see that $k$ is maximum when $y=-x+k$ contacts with the circle from above.
The distance from the center of the above circle from the linear graph $d=\dfrac{|-1-k|}{\sqrt{1^2+1^2}}$ should be the same with the radius of the circle( $6$ ).
$|-1-k|=\sqrt{2}\times6$
And since $k>0,$
$k=-1+6\sqrt{2}=p+q\sqrt{2}.$
Therefore $p=-1,~q=6.$