Question

Find the equation of the circle with center that passes through point A.
center (1, 3) and point A (4, -2) equaling to 0

Answer 1

whats the equation of a generic circle?

Answer 2

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

Answer 3

Use the distance formula to find the radius

Answer 4

good, now replace h with 1 and k with 3 for starters

Answer 5

the radius "r" is just the distance from the center to A

Answer 6

(x-1)^2+(y-3)^2=r^2 .....

Answer 7

good so far; now for "r"

Answer 8

do you recall the formula or any method to find the distance between 2 points?

Answer 9

yeahhhhh

Answer 10

then plug in the point for center and the point for A to determine the distance; thats your r then

Answer 11

i get 3 over and 5 down sooo;

sqrt(3^2+5^2) = sqrt(34) perhaps?

Answer 12

and r^2 is then 34

(x-1)^2 + (y-3)^2 = 34

expand it all out and put all things to one side to zero it out

Answer 13

wait im not sure how to zero that out..

Answer 14

expand your ^2 terms

Answer 15

whats (x-1)^2 ?

Answer 16

x^2-1^2

Answer 17

not quite,

(x-1)(x-1) = x^2 -2x +1

Answer 18

expand the (y-3)^2

Answer 19

y^2-6^2+3

Answer 20

.... youre really bad at that :)

y-3
y-3
----
y^2 -3y
-3y+9
----------
y^2 -6y + 9

Answer 21

sooo:

x^2 -2x +1 + y^2 -6y +9 = 34

combine the constants

x^2 -2x + y^2 -6y +10 = 34

and -34 from each side

x^2 -2x + y^2 -6y +10 = 34
-34 -34
---------------------------
x^2 -2x + y^2 -6y - 24 = 0

Answer 22

gochaaa , thanks!

Answer 23

good luck :)