what do you do to find the equation of a circle with diameters at end points (5,13) and (2,-8)?

Answer 1

Is that a single diameter? Your statement is ambiguous.

Assuming that it is, find the distance between those two points. Halfway between the two points is the center

Answer 2

Using section formula, find co-ordinates of centre.
let the centre be C(h,k)

h=(5+2)/2=7/2
k=(13-8)/2=5/2

Now find the radius(r) using distance formula,

r^2 = (2-7/2) ^2 + (-8-5/2)^2
=>r^2= (-3/2)^2 + (-21/2)^2
=>r^2=9/4 + 441/4=450/4=225/2

let there be a point (x,y) on the circumference of the circle.

=>(x-7/2)^2 + (y-5/2)^2 = 225/2
=>x^2 + 49/4 -7x + y^2 +25/4 -5y=225/2

=>4x^2 + 4y^2 -28x - 20y +74=450
=>4x^2 + 4y^2 -28x - 20y +74-450=0
divide both sides by 4
therefore, eq. is x^2 + y^2 -7x - 5y -94=0

Answer 3

Center is at the midpoint of the two endpoints which is ( (5 + 2) / 2, (13 - 8) / 2) = (7/2, 5/2).

Radius will be half of the distance between the two original points:

r = 1/2 sqrt[ (5-2)^2 + (13 + 8) ^2 ] = 1/2 sqrt[450]

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

(x - 7/2)^2 + (y - 5/2)^2 = 225/2

Answer 4

If A(x1,y1) and B(x2,y2) are the extremities of a diameter of a circle , the equation of the circle can be given by the formula:
(x-x1) (x-x2) + (y-y1) (y-y2) =0.
Thus, the eqn of the circle whose diameters have end points
(5,13) and (2,-8) will be:
(x-5) (x-2) + (y-13) (y-(-8)) =0
on solving further we get:
x^2 - 7x +10 + y^2 - 5y -104 = 0
i.e x^2 + y^2 -7x -5y -94 =0

Answer 5

endpoints of the diameter are (-4,4) and (2,8) for the middle use the midpoint formula midpoint formula: Xm=(X?+X?)/2 Ym=(Y?+Y?)/2 X=(-4+2)/2 =>X=-a million Y=(4+8)/2 =>Y=6 the middle is (-a million,6) use now distance formula for the size of the diameter: d²=(x?-x?)²+(y?-y?)² d²=(2-(-4))²+(8-4)² =>d²=fifty two =>d=?fifty two =>d=2?13 the diameter is as two times by way of fact the radius so the radius is ?13 the equation of the circle is: (x+a million)²+(y-6)²=13

Answer 6

Ok, you find the radius by doing distance formula then divide it by two. Then you find the midpoint (midpt formula) and plug it into the formula which is
(x-h)^2 + (y-k)^2 = r^2

Answer 7

ist find the diameter
sqrt((13--8)^2+(2-5)^2)=sqrt(21^2+3^2=sqrt(450)=15sqrt(2)
so the radius is 7.5sqrt(2)

next find the center

(2+5)/2=3.5
(13-8)/2=2.5

so the center is at (3.5,2.5)
equation is
(x-3.5)^2+(y-2.5)^2=112.5

Answer 8

i don't know. but got opportunity to learn from you Q thank you