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Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 10"

(Created page with "==Problem== Circle <math>\omega_1</math> is defined by the equation <math>(x-7)^2+(y-1)^2=k,</math> where <math>k</math> is a positive real number. Circle <math>\omega_2</math> ...")
 
 
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From the distance formula, <math>(C_1C_2)^2=R^2=(x-7)^2+(27-7x)^2=50x^2-392x+778</math>. Setting the previous two equations equal to each other yields <math>R^2=50x^2-392x+734+r^2=50x^2-392x+778</math>. The <math>x</math> terms nicely cancel out, leaving <math>734+r^2=778</math>. Thus <math>r^2=k=\boxed{044}</math>.
 
From the distance formula, <math>(C_1C_2)^2=R^2=(x-7)^2+(27-7x)^2=50x^2-392x+778</math>. Setting the previous two equations equal to each other yields <math>R^2=50x^2-392x+734+r^2=50x^2-392x+778</math>. The <math>x</math> terms nicely cancel out, leaving <math>734+r^2=778</math>. Thus <math>r^2=k=\boxed{044}</math>.
 +
 +
==Video Solution by Punxsutawney Phil==
 +
https://youtube.com/watch?v=U_KqWvheyEk

Latest revision as of 21:28, 26 December 2022

Problem

Circle $\omega_1$ is defined by the equation $(x-7)^2+(y-1)^2=k,$ where $k$ is a positive real number. Circle $\omega_2$ passes through the center of $\omega_1$ and its center lies on the line $7x+y=28.$ Suppose that one of the tangent lines from the origin to circles $\omega_1$ and $\omega_2$ meets $\omega_1$ and $\omega_2$ at $A_1,A_2$ respectively, that $OA_1=OA_2,$ where $O$ is the origin, and that the radius of $\omega_2$ is $\frac{2011} {211}$. What is $k$?

Solution

Let the centers of $\omega_1,\omega_2$ be $C_1,C_2$ and let their radii be $r,R$ respectively. From the given information, $C_1=(7,1)$ and $C_2=(x,28-7x)$ for some $x$. Using the distance formula between $O$ and $C_1$ yields $C_1O=\sqrt{(7-0)^2+(1-0)^2}=\sqrt{50}$. From right triangle $\Delta C_1A_1O$, we have $r^2+(A_1C)^2= 50$. Rearranging yields $A_1C=\sqrt{50-r^2}$.


Similarly, using the distance formula between $O$ and $C_2$ yields $C_2O=\sqrt{x^2+(28-7x)^2}$. From right triangle $\Delta C_2A_2O$, we have $R^2+(A_2C)^2= x^2+(28-7x)^2$. Substituting for $A_2C$ and rearranging yields $R^2=x^2+(28-7x)^2-50+r^2=50x^2-392x+734+r^2$.


From the distance formula, $(C_1C_2)^2=R^2=(x-7)^2+(27-7x)^2=50x^2-392x+778$. Setting the previous two equations equal to each other yields $R^2=50x^2-392x+734+r^2=50x^2-392x+778$. The $x$ terms nicely cancel out, leaving $734+r^2=778$. Thus $r^2=k=\boxed{044}$.

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=U_KqWvheyEk

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