Draw a circle with center O.
Line AD is a chord that is 8cm long. The arc above is smaller than the one below.
B is the center of AD.
Line CB is a line that is 2cm long. It meets AD at 90°.
Diagram:
Given the facts above, is it possible to find CO (the radius of the circle)? If so, what is the radius, and how can I find it for any other circle?
$\endgroup$ 55 Answers
$\begingroup$Let $CO=r$ then we have $r^2=(r-2)^2+4^2$. Solve for $r=5$
$\endgroup$ 6 $\begingroup$Hint: Suppose $OB=x$, then $OA=OC=x+2$. Now apply Pythagoras theorem in triangle $AOB$
$\endgroup$ $\begingroup$Because $D$ lies on the circle, $OD^2=OB^2+BD^2$. Also because $OD(=OC)=OB+BC$, $OB=OD-BC$. Substituting $OB$ in the first equation yields
$ $ $ $ $ $ $ $ $OD^2=(OD-BC)^2+BD^2$
$≡$ $OD^2= OD^2-2\centerdot OD\centerdot BC+ BC^2+BD^2$
$≡$ $2\centerdot OD\centerdot BC =BC^2+BD^2$
$≡$ $OD = (BC^2 + BD^2)/(2\centerdot BC)$
For the example at hand, this results in a radius of $(2^2+4^2)/(2\centerdot 2) = 5$.
$\endgroup$ $\begingroup$Let $\overline{CC^\prime}$ be a diameter.
By Power of a Point, $AB \cdot BD = CB \cdot BC^\prime = 4 \cdot 4 = 2 \cdot BC^\prime$.
Then $BC^\prime = 8$, $CC^\prime = 10$, and $CO = 5$.
$\endgroup$ 0 $\begingroup$Edited your sketch. When two line segments cut at B, use circle segments property of constant products:
$ BA . BD = BC. BE ; 2 ( 2 R -2) = 4^2 $ so $ R = 5. $
