Geometry #1

by sjaelee, Aug 16, 2011, 9:59 AM

Quote:
A machine-ship cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of AB is 6 cm and that of BC is 2 cm. The angle ABC is a right angle. Find the square of the distance (in centimeters) from to the center of the circle.

Source: AIME

We make the center of the circle the origin.

Then we have the equation $x^2+y^2=50$. Note that both $A$ and $C$ are on

the circle, so they satisfy the equation. And C is $6$ down, $2$ right of A.

Let $A=(x,y)$.

$x^2+y^2=50$

$(x+2)^2+(y-6)^2=50$.

That means $4x+4-12y+36=0$, so $x-3y+10=0$. That means $x=3y-10$.

Aha! We can substitute.

$(3y-10)^2+y^2=50$, so

$10y^2-60y+100=50$,

$10y^2-60y+50=0$. Using qudratic formula,

$\frac{60\pm\sqrt{3600-4(10)(50)}}{20}$, so

$y=5,1$. We use $5$ becuase one is a reflection of what we have (same).

$A=(5,5)$ and $C=(-1,7)$. Point $B$ is simply two left of $C$, so

$B=(-1,5)$. The distance between that and the centter, the origin, is

$\sqrt{(-1)^2+(5)^2}=\sqrt{26}$. Squaring,

$\sqrt{26}^2=26$. Thus,

$26$.
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