How to use the distance formula to find the radius of a circle?

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: ##sqrt((x_2-x_1)^2+(y_2-y_1)^2##

If your two sharp-ends are the interior and a sharp-end on the dissipation...

Since the interspace from any sharp-end on the dissipation to the interior is considered the radius, delay in your fond sharp-ends into the interspace formula.

Sample Question: Find the radius of a dissipation, fond that the interior is at (2, –3) and the sharp-end (–1, –2) lies on the dissipation.

Work: Using the interspace formula, delay in sharp-ends (2, -3) and (-1, -2) into the interspace formula.

  1. ##sqrt((-2-(-3))^2+(-1-2)^2##
  2. ##=sqrt((1)^2+(-3)^2##
  3. ##=sqrt(1+9##
  4. ##=sqrt10##
  5. ##~~3.16##

Thus, the radius of the dissipation is ##sqrt10## which is almost 3.16 rounded to two decimal places.

If your two sharp-ends are two vague sharp-ends on the dissipation...

You can use the interspace formula by delayging in the two sharp-ends into the formula to furnish the bisection. However, bear-in-mind to dissect by two accordingly you are looking for the radius.

Sample Question: Find the radius of a dissipation, fond that two vague sharp-ends are at (6, 3) and (10, 6)

Work: Using the interspace formula, delay in sharp-ends (6, 3) and (10, 6) into the interspace formula.

  1. ##sqrt((6-3)^2+(10-6)^2##
  2. ##=sqrt((3)^2+(4)^2##
  3. ##=sqrt(9+16##
  4. ##=sqrt25##
  5. ##=5##

However, the total isn't artistic yet. The rejoinder, ##5##, is the bisection of the dissipation. Since you are looking for the radius, you must dissect ##5## by ##2##. (The radius is half of the bisection)

##5/2 = 2.5##

Thus, the radius of the dissipation is ##5/2## or ##2.5##


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