circle confusion

[SpGcAr1117]

Okay, well, I'm in geometry and we're learning about inscribed angles and such and I understand the majority of it but there are a couple of problems I just don't get how to do.

In an attempt to [hopefully] make getting an answer easier (and quicker?), I've made this image of the two problems I need help with:

We have to find the radius of each circle.

Please help.
[/img]

 

[Denis]

#11: join centre to one of the ends of the chord;
you now have a right triangle with hypotenuse = r and legs = 10 and r-8

#12: join centre to one of the ends of the chord;
you now have a right triangle with hypotenuse = r and legs = 6 and 10-r

 

[soroban]

Hello, SpGcAr1117!

These have little to do with inscribed angles ... mostly Pythagorus.

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11.
              * * *
          *           *
        *               *
       *                 *

      *         O         *
      *         *         *
      *         | \       *
            /   |   \ r
       *  / 10  |  10 \  *
      A * - - - * - - - * B
          *    D|8    *
              * * *
                C

We have a circle with center $\displaystyle O.$
Radii $\displaystyle OA\,=\,OB\,=\,OC\,=\,r$

Radius $\displaystyle OC$ bisects chord $\displaystyle AB$ at $\displaystyle D$
$\displaystyle \;\;$and $\displaystyle OC\,\perp\,AB$

We are told that: $\displaystyle \,DC\,=\,8,\;AD\,=\,DB\,=\,10$

We see that: $\displaystyle \,OD\,=\,r\,-\,8$

In right triangle $\displaystyle ODB:\;(r\,-\,8)^2\,+\,10^2\:=\:r^2$

We have: $\displaystyle \,r^2\,-\,16r\,+\,64\,+\,100\:=\:r^2\;\;\Rightarrow\;\;16r\,=\,36\;\;\Rightarrow\;\;r\,=\,\frac{9}{4}$

12.
                C
              * * *
          *  6  |  6  *
      A *       +       * B
       *  \     |D   /  *
            \   |   / r
      *       \ | /       *
      *         *         *
      *         |O        *
                |
       *        |        *
        *       |       *
          *     |     *
              * * *
                E

We have a circle with center $\displaystyle O.$
Radii $\displaystyle OA\,=\,OB\,=,OC\,=\,OE\,=\,r.$

$\displaystyle OC$ bisects chord $\displaystyle AB$ at $\displaystyle D$
$\displaystyle \;\;$and $\displaystyle OC\,\perp\,AB.$

We are told that $\displaystyle DE\,=\,10,\;AD\,=\,DB\,=\,6$

We see that: $\displaystyle \,DO\,=\,10\,-\,r$

In right triangle $\displaystyle ODB:\;(10\,-\,r)^2\,+\,6^2\;=\;r^2$

We have: $\displaystyle \,100\,-\,20r\,+\,r^2\,+\,36\;=\;r^2\;\;\Rightarrow\;\;20r\,=\,136\;\;\Rightarrow\;\; r\,=\,\frac{34}{5}$


Edit: Too fast for me, Denis!