Question

Write the standard form of the equation of a circle with endpoint of a diameter at the points (9,6) and (-5,9)

Answer 1

find the radius using distance formula and dividing that answer by 2, then use x^2 + Y^2= R^2

Answer 2

R being the radius!

Answer 3

I have (x-5)squared + (y-15/2) squared - but I can't get the last part - what the radius is

Answer 4

use distance formula.. to find diameter.. then divide by 2

Answer 5

do you know what distance formula is?

Answer 6

ugh, no

Answer 7

Is it 10 or 100?

Answer 8

i ll tell distance formula (x1 y1) and (x2 y2).. distance between them is
\[\sqrt{(x1-x2)^{2}+(y1-y2)^{2}}\]

Answer 9

174?

Answer 10

No do \[\sqrt{(9+5)^{2}+(6-9)^{2}}\]

Answer 11

\[\sqrt{205}\]?

Answer 12

correct.. !

Answer 13

so the answer is: (x+5)\[(x-5)^{2}+(y-15/2)^{2}=205?\]+(y-@DIV{15;2})@Sup{2}=205?

Answer 14

that doesn't look right

Answer 15

Noo.. what you got is the diameter.. Root(205)/2 is the radius.. so answer is ............ = 205/4

Answer 16

sorry, I am more confused than ever

Answer 17

how do I write the standard equation?

Answer 18

(x-a)squared +(y-b)squared= (radius)squared

Answer 19

you first tell me what is the co ordinate of the centre of the circle, ??? those co ordinates are (a, b).. what you got i believe is not correct

Answer 20

2, 15/2