Simple Circle?

Question

Find an equation of the circle that satisfies the conditions:

endpoints of the diameter A ( 16, - 5) and B ( - 10, 7)

Express your answer in standard form of a circle.

Answer 1

Okely dokely.
There are squillions of standard formulae for various types of circles, you can get them at:
http://mathworld.wolfram.com/Circle.html

Here, the circle S given P(x1,y1)Q(x2,y2) as a diameter is given by
(x-x1)(x-x2) + (y-y1)(y-y2) = 0

Thus:
(x-16)(x+10) + (y+5)(y-7) = 0
x²-6x -160 + y²-2y -35 = 0
(x-3)² + (y-1)² = 195+9+1 ; completing the squares on LHS and adding corresponding amounts 3² and 1² on RHS....
(x-3)² + (y-1)² = 205
(x-3)² + (y-1)² = (√205)²


\What james did is cleverer and faster but if it's for an exam I think they would expect you to use the standard formulae.

Answer 2

Generally, to write an equation for a circle, you need to find its center and its radius so that you can use the standard form: (x-a)^2+(y-b)^2 = r^2, where (a,b) is the coordinates of the center, and r is the radius.

Applying midpoint theorem and distance formula, you can write the equation directly,

[x-(16-10)/2]^2+[y-(-5+7)/2]^2
= [(16+10)^2+(7+5)^2]/4......(1)
Reason: r^2 = d^2/4,where d is the diameter.

Simplifying (1) leads to,
(x-3)^2+(y-1)^2 = 205

Answer 3

the definition of a circle with center (a,b) and radius r is

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


first you need the center between the two lines.
this is equal to the mid point...which is the
average of the two points.

(16+-10)/2=x=6/2=3
(-5+7)/2=y=2/2=1

so mid point =(3,1)

next you need R the radius
this is the distance of the midpoint to one of
the points. Use distance formula

sqrt((16-3)^2+(-5-1)^2)=
sqrt(13^2+-6^2)=
sqrt(205)

so the answer is
(x-3)^2+(y-1)^2=sqrt(205)^2

Answer 4

The center would be at the diameter midpoint: (3,1). The radius is half of sqrt( 26^2+12^2)=14.317=sqrt(205). Therefore, the equation would be: (x-3)^2+(y-1)^2=205.

Answer 5

a=Pi R squared