Question

If the endpoints of the diameter of a circle are (−6, 6) and (6, −2), what is the standard form equation of the circle?

Answer 1

Answer:

Hence, equation of circle is:

$x^2+y^2-4y=48$

Step-by-step explanation:

Te end points of the diameter are given to be:

(−6, 6) and (6, −2).

We know that the center of the circle is the mid-point of the diameter.

and the coordinates of the mid-point (e,f) of (a,b) and (c,d) is calculated as:

$e=\dfrac{a+c}{2}, f=\dfrac{b+d}{2}$

Here we have (a,b)=(-6,6) and (c,d)=(6,-2).

Hence, the coordinates of center (e,f) is calculated as:

$e=\dfrac{-6+6}{2},f=\dfrac{6-2}{2}\\\\e=\dfrac{0}{2},f=\dfrac{4}{2}\\\\e=0,f=2$

Hence, center is located at: (0,2)

Now, the radius of circle is the the distance between the center and a point on the circle.

i.e. distance between (0,2) and (6,-2).

We know that the distance between two points (a,b) and (c,d) is calculated as:

$=\sqrt{(c-a)^2+(d-b)^2}$

Here we have (a,b)=(0,2) and (c,d)=(6,-2)

Hence, the length of the radius is calculated as:

$\sqrt{(6-0)^2+(-2-2)^2}\\\\=\sqrt{36+16}\\\\=\sqrt{52}\\\\=2\sqrt{13}$

Hence, the equation of the circle with center (h,k) and radius r is given by:

$(x-h)^2+(y-k)^2=r^2$

Hence, the equation of circle is:

$(x-0)^2+(y-2)^2=(2\sqrt{13})^2\\\\x^2+y^2+4-4y=52\\\\x^2+y^2-4y=48$

Hence, equation of circle is:

$x^2+y^2-4y=48$

Answer 2

i  Acehello :
let : A(-6,6)  B(6,-2)
the center is w((-6+6)/2 , (6-2)/2)....(midel [ AB]
w(0 ,2)
the ridus is : r = AB/2
AB = √(-6-6)²+(6+2)² = √(144+64) =√208/2
an equation in standard form equation of the circle :
(x+0)²+(y-2)² =  (√208/2)²=208/4 = 52