Question #0994c

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

`AC^2= OC^2 + OA^2`

`25^2= 7^2+ OA^2`

`OA = sqrt(625 - 49)`

`OA = 24`

The coordinates of point `A = (0,24)`

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

![www.desmos.com/calculator]()

Please observe that, the points A, B, and C are in black, the sides of `DeltaABC` 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= (7,0)` to point `A = (0,24)`:

`m = (24-0)/(0-7)`

`m = -24/7`

The slope, n, of its perpendicular bisector is:

`n = -1/m`

`n = -1/(-24/7)`

`n = 7/24`

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

`((0+7)/2, (24+0)/2) = (3.5,12)`

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

`y = 7/24(x - 3.5)+12`

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

![www.desmos.com/calculator]()

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

`y = 7/24(0 - 3.5)+12`

`y = 527/48`

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

`r = 24 - 527/48`

`r = 625/48" cm"`

Here is a graph with the circle added.

![www.desmos.com/calculator]()