Mock Geometry AIME 2011 Problems/Problem 10

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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Mock Geometry AIME 2011 Problems/Problem 10

Problem

Solution

Video Solution by Punxsutawney Phil