# 2012 AMC 12B Problems/Problem 10

## Problem

What is the area of the polygon whose vertices are the points of intersection of the curves $x^2 + y^2 =25$ and $(x-4)^2 + 9y^2 = 81 ?$ $\textbf{(A)}\ 24\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 37.5\qquad\textbf{(E)}\ 42$

## Solution 1

The first curve is a circle with radius $5$ centered at the origin, and the second curve is an ellipse with center $(4,0)$ and end points of $(-5,0)$ and $(13,0)$. Finding points of intersection, we get $(-5,0)$, $(4,3)$, and $(4,-3)$, forming a triangle with height of $9$ and base of $6.$ So the area of this triangle is $9 \cdot 6 \cdot 0.5 =27 \textbf{ (B)}.$

## Solution 2

Given the equations $x^2 + y^2 =25$ and $(x-4)^2 + 9y^2 = 81$, we can substitute $y^2=25-x^2$ from the first equation and plug it in to the 2nd equation, giving us $(x-4)^2+9(25-x^2)=81$. After rearranging, $8x^2+8x-160=0$ or $x^2+x-20=0$. The solutions are $x=-5$ and $x=4$. This gives us the points $(-5,0),(4,3)$,and $(4,-3)$. The area of the triangle formed by these points is $27=\fbox{B}$ ~dragnin (minor edit by KevinChen_Yay)

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