Suppose that we have a right triangle $abc$ with the right angle at $b$ such that $ac = \sqrt{61}$ and $ab = 5.$ a circle is drawn with its center on $ab$ such that the circle is tangent to $ac$ and $bc.$ if $p$ is the point where the circle and side $ac$ meet, then what is $cp$?

  CP = 6

Step-by-step explanation:

The length of segment BC is given by the Pythagorean theorem:

  AC² = AB² +BC²

  (√61)² = 5² + BC² . . . . . fill in the given numbers

  61 -25 = BC² = 36 . . . . .subtract 25

  BC = 6 . . . . . . . . . . . . . . take the square root

Since the center of the circle is on AB and is tangent to BC, it must pass through point B. That is, segment BC of length 6 is one of the tangent lines from point C. The other one, to point P, must be the same length, so ...

  CP = 6

here is a big fat