Find the equation of the circle, in standard form, where (1, –5) and (–3, 6) are end points of a diameter.

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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We know that the equation of the circle into the standard form is:

(x-a)^2 + (y-b)^2 = r^2  where (a,b) is the center is the circle and r is the radius.

Given the points (1,-5) and (-3,6) are the end points of the diameter.

Then the midpoint will be the center of the circle.

==> Mx = (1-3)/2 = -2/2 = -1

==> My = (-5+6)/2 = 1/2

Then the center (a,b) = (-1, 1/2)

==> (x+1)^2 + (y-1/2)^2 = r^2

Now let us calculate the radius.

First, we can calculate the length of the diameter by using the distance between two points.

==> D = sqrt[(1+3)^2+ (-5-6)^2]

           = sqrt(16+121) = sqrt(137)

Then the diagonal is sqrt137 units.

Then we know that the radius (r) = D/2 = sqrt137/ 2

Now we will substitute into the equation.

==> (x+1)^2 + (y-1/2)^2 = (sqrt137/2)^2

==> x^2 +2x+1 + y^2 -y + 1/4 = 137/4

==> x^2 + y^2 + 2x -y = 137/4 - 5/4

==> x^2 + y^2 + 2x -y  = 132/4

==> x^2 + y^2 + 2x -y = 33

Then, the equation of the line is given by:

==> x^2 + y^2 + 2x -y -33 = 0

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to find the circle which has ( 1, -5) and ( -3 , 6) as the end points of a diameter.

Now, the distance between the two points is  sqrt [( 1 +3)^2 + (-5-6)^2]

=> sqrt [ 4^2 + 11^2]

=> sqrt ( 16 + 121)

=> sqrt 137.

The radius is (sqrt 137)/2.

The mid point between the two point given is ( -2/2 , 1/2)

The equation of the circle is ( x +1)^2 + (y - 1/2)^2 = (sqrt 137)^/4

=> x^2 + 2x + 1 + y^2 + 1/4 - y = 137 / 4

=> x^2 + y^2 + 2x - y + 5/4 = 137 / 4

=> x^2 + y^2 + 2x -y = 132 / 4

=> => x^2 + y^2 + 2x -y = 33

=> x^2 + y^2 + 2x -y -33 = 0

The equation of the circle is x^2 + y^2 + 2x + -y -33 = 0

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