Olympiad Maths Prep

Track / Stage 3 / 135 of 260 #135 of 2000

Problem 135

AMC 10/12, early questions
Geometry Difficulty 3.5 Find the answer

A circle has center (−10,−4)(-10, -4) and has radius 1313. Another circle has center (3,9)(3, 9) and radius 65\sqrt{65}. The line passing through the two points of intersection of the two circles has equation x+y=cx+y=c. What is cc?
(A) 3(B) 33(C) 42(D) 6(E) 132\textbf{(A)}\ 3\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{13}{2}

Official solution

The equations of the two circles are (x+10)2+(y+4)2=169(x+10)^2+(y+4)^2=169 and (x−3)2+(y−9)2=65(x-3)^2+(y-9)^2=65. Rearrange them to (x+10)2+(y+4)2−169=0(x+10)^2+(y+4)^2-169=0 and (x−3)2+(y−9)2−65=0(x-3)^2+(y-9)^2-65=0, respectively. Their intersection points are where these two equations gain equality. The two points lie on the line with the equation (x+10)2+(y+4)2−169=(x−3)2+(y−9)2−65(x+10)^2+(y+4)^2-169=(x-3)^2+(y-9)^2-65. We can simplify this like the following. (x+10)2+(y+4)2−169=(x−3)2+(y−9)2−65→(x2+20x+100)+(y2+8y+16)−(x2−6x+9)−(y2−18y+81)=104→26x+26y+26=104→26x+26y=78→x+y=3(x+10)^2+(y+4)^2-169=(x-3)^2+(y-9)^2-65 \rightarrow (x^2+20x+100)+(y^2+8y+16)-(x^2-6x+9)-(y^2-18y+81)=104 \rightarrow 26x+26y+26=104 \rightarrow 26x+26y=78 \rightarrow x+y=3. Thus, c=(A) 3c = \boxed{\textbf{(A)}\ 3}.
Solution by TheUltimate123

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.