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

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

Mathematics
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whats the equation of a generic circle?
(x-h)^2+(y-k)^2=r^2
Use the distance formula to find the radius

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good, now replace h with 1 and k with 3 for starters
the radius "r" is just the distance from the center to A
(x-1)^2+(y-3)^2=r^2 .....
good so far; now for "r"
do you recall the formula or any method to find the distance between 2 points?
yeahhhhh
then plug in the point for center and the point for A to determine the distance; thats your r then
i get 3 over and 5 down sooo; sqrt(3^2+5^2) = sqrt(34) perhaps?
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
wait im not sure how to zero that out..
expand your ^2 terms
whats (x-1)^2 ?
x^2-1^2
not quite, (x-1)(x-1) = x^2 -2x +1
expand the (y-3)^2
y^2-6^2+3
.... youre really bad at that :) y-3 y-3 ---- y^2 -3y -3y+9 ---------- y^2 -6y + 9
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
gochaaa , thanks!
good luck :)

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