Question

PLEASE HELP!!!! The edges of the diameter of a circle are at (-3,-8) and (5,7).

Answer 1

Answer:

  x^2 + y^2 -2x +y -71 = 0

Step-by-step explanation:

The center of the circle is the midpoint of the diameter, so is ...

  ((-3, -8) +(5, 7))/2 = (-3+5, -8+7)/2 = (1, -1/2)

The square of the radius of the circle can be found from the Pythagorean theorem. The x- and y- differences between an end point and the center are the legs of a right triangle with r as the hypotenuse.

  r^2 = (5 -1)^2 +(7 -(-1/2))^2 = 4^2 +7.5^2 = 16 +56.25

  r^2 = 72.25

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The standard form equation of the circle with center (h, k) and radius r is ...

  (x -h)^2 +(y -k)^2 = r^2

  (x -1)^2 +(y -(-1/2))^2 = 72.25

Eliminating parentheses, we have ...

  x^2 -2x +1 +y^2 +y +0.25 = 72.25

Subtracting the right-side constant and rearranging to descending powers, we find the general form of the equation to be ...

  x^2 + y^2 -2x +y -71 = 0

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Comment on the work

Of course, you know that ...

  (a -b)^2 = a^2 -2ab +b^2

This helps you simplify the squares of binomials.