write an equation in standard form of the circle with diameter endpoints (2,-3) and (6,-11)?

Question

please help me i dont know how to solve this question!!!

Answer 1

first you need to find the center (h,k) by finding the midpoint of the diameter.then find the radius(r) of the circle by distance formula..supply the (h,k) and r to the equation
(x-h)^2 + (y-k)^2=r^2

find (h,k).

midpoint = {( x2+x1)/2, (y2+y1)/2}
= {(2+6)/2, (-3+ -11)/2}
= (4, -7)
find r
distance = square root of {(y2-y1)^2 + (x2-x1)^2}
= sq.rt of {[-3 - (-11)]^2 + (2-6)^2}
= sq.rt of (64 + 16)
= sq rt of 80

hence (x-4)^2 + ((y-(-7))^2 = (sq rt of 80)^2
x^2 -8x +16 +y^2 +14y + 49 = 80
x^2 + y^2 -8x + 14y = 80 - 16 - 49
x^2 + y^2 -8x + 14y = 15 //answer

Answer 2

(x-4)^2 + (y -{-7})^2 = [2times sqrt 5]^2 or 20

center (4, -7) radius 2 times sqrt 5

[GUNRUNNY: Let me add/suggest that you think of the mid pt formula as (average of the x's of the 2 pts, average of the y's of the 2 pts)

In the distance formula (between 2 pts) it doesn't matter whether you do (z1 - z2) squared or (z2 - z1) squared; the important thing is that you take a difference between the (in this illustration) z coordinates of the 2 pts. That difference squared will be the same regardless of the sign of the difference. ]

Finally, 13angus13 introduces an error in his work at line: hence... where his computed DIAMETER is presumed to be a radius.

Answer 3

This is pretty easy if you have an equation chart. You don't have to solve for anything except the center. Then you just have to write an equation for the circle in standard form. I don't have mine with me now...but just look it up.

Answer 4

use the midpoint formula to find the center of the radius. its gonna be a coordinate pair like (2,3) as an ex.
then use the distance formula to find the distance between the given points divide that by 2 and thats ur radius and then just plug it into the standard formula for a circle

Answer 5

♠ the distance btw points is diameter, the center being the midpoint
c=(p1+p2)/2 = (2*i –3*j +6*i –11*j)/2 = (8*i–14*j)/2 =(4,-7);
♣ R=diameter/2 = 0.5*sqrt(p1x-p2x)^2 +(p1y-p2y)^2) =
=0.5*sqrt((2-6)^2 +(-3+11)^2) = 0.5*sqrt(16+64) = 2sqrt5;
♦ thus (x-4)^2 +(y+7)^2=20;