# Question #0994c

Nov 7, 2017

$r = \frac{625}{48} \text{ cm}$

#### Explanation:

Orient the triangle so that BC, is on the x axis, point B is $\left(- 7 , 0\right)$, and point C is $\left(7 , 0\right)$; this makes the y axis the perpendicular bisector of chord BC at the origin $O = \left(0 , 0\right)$. Because the y axis forms a right triangle, $\Delta O A C$, with side AC as the hypotenuse, we can use the Pythagorean Theorem to find the coordinates of point A:

$A {C}^{2} = O {C}^{2} + O {A}^{2}$

${25}^{2} = {7}^{2} + O {A}^{2}$

$O A = \sqrt{625 - 49}$

$O A = 24$

The coordinates of point $A = \left(0 , 24\right)$

Here is a graph of what I have described thus far:

Please observe that, the points A, B, and C are in black, the sides of $\Delta A B C$ are in green, and the perpendicular bisector of side BC is in purple.

The slope of the line AC is the slope from point $C = \left(7 , 0\right)$ to point $A = \left(0 , 24\right)$:

$m = \frac{24 - 0}{0 - 7}$

$m = - \frac{24}{7}$

The slope, n, of its perpendicular bisector is:

$n = - \frac{1}{m}$

$n = - \frac{1}{- \frac{24}{7}}$

$n = \frac{7}{24}$

The perpendicular bisector will go through the midpoint between A and C:

$\left(\frac{0 + 7}{2} , \frac{24 + 0}{2}\right) = \left(3.5 , 12\right)$

The point-slope form of the equation of the perpendicular bisector is:

$y = \frac{7}{24} \left(x - 3.5\right) + 12$

Here is a graph of the triangle with the perpendicular bisector:

The center of the circle is the point where this line intercepts the y axis:

$y = \frac{7}{24} \left(0 - 3.5\right) + 12$

$y = \frac{527}{48}$

The radius of the circle is the distance from the y intercept to point A

$r = 24 - \frac{527}{48}$

$r = \frac{625}{48} \text{ cm}$

Here is a graph with the circle added.