2024 DMC Mock 10 Problems/Problem 16

Problem

Let $(x,y)$ be real numbers which satisfy $(x-48)^2+(y-14)^2=169$ . What is the maximum possible value of $\sqrt{x^2+y^2}$ ?

$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 37\qquad\textbf{(C)}\ 50\qquad\textbf{(D)}\ 63\qquad\textbf{(E)}\ 65$

Solution

The maximum possible distance from the center to a point on the circle (which is what the question is asking for) is the point such that that point, the circle center, and the origin are all collinear. Thus the distance from the origin to the circle center is $\sqrt{14^2+48^2}=50$ by the Pythagorean Theorem, and the radius is $13$ , so the maximum is $50+13=\boxed{\textbf{(D)}\ 63}$ .

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2024 DMC Mock 10 Problems/Problem 16

Problem

Solution