For each point, compute $x-(-p)$ and $y-q$ (equivalently $x+p$ and $y-q$).

A correct option should make $x+p$ and $y-q$ constant numbers so that their squares can add to 1369.

Enter the circle as (x+p)^2+(y-q)^2=1369. If Desmos asks, create sliders for p and q.

Set p=1 and q=2 (any values work). The circle will update. ©‍ aniко.аi

Enter the four points from the choices (using the same p and q). For example, one point would be (-p+35, q+12).

See which plotted point lies exactly on the circle. Then change p and q to new values (like p=3, q=-1) and observe which choice still lies on the circle. The one that stays on the circle is the correct answer.

The equation

has center $(-p, q)$ and radius $\sqrt{1369}=37$.

A point $(x,y)$ is on the circle exactly when the distance from $(-p,q)$ is 37: © аnікo.‌aі

So we just need $(x+p)$ and $(y-q)$ to be numbers whose squares add to 1369.

Compute $(x+p)$ and $(y-q)$ for each option.

• For $(p+12, q+35)$, $x+p=(p+12)+p=2p+12$, which depends on $p$, so it is not guaranteed to equal 37.

• For $(-p+12, -q-35)$, $y-q=(-q-35)-q=-2q-35$, which depends on $q$.

• For $(p-35, -q+12)$, both $x+p=(p-35)+p=2p-35$ and $y-q=(-q+12)-q=-2q+12$ depend on $p$ or $q$.

For $(-p+35, q+12)$:

• $x+p=(-p+35)+p=35$
• $y-q=(q+12)-q=12$

Then

so this point lies on the circle. Therefore, the correct choice is $(-p + 35, q + 12)$.