Question

ook now this one find the radius of the circle defined by the equation x^+y^2-10x+4y+4=0

Answer 1

i found it i got a different one

Answer 2

Complete the square to first get the equation of the circle.
x^2+y^2-10x+4y+4=0 (You forgot a power 2 somewhere there)
(x-5)^2 - 5^2 + (y+2)^2 -2^2 + 4 = 0
(x-5)^2 + (y+2)^2 = 25-4 + 4
(x-5)^2 + (y+2)^2 = 25
Centre is thus (5,-2) and the radius, which you have already found, should be sqrt (25) = 5 units

Answer 3

\[x ^{2}-10x+y ^{2}+4y+4=0\]
\[x ^{2}-10x+25+y ^{2}+4y+4=25\]
\[(x-5)^{2}+(y+2)^{2}=25\]

Answer 4

ook now this one find the radius of the circle defined by the equation 3x^2+3y^2+42x+42y+186=0

Answer 5

divide 3 on both sides firstly

Answer 6

um hw

Answer 7

\[(3x ^{2}+3y^{2}+42x+42y+186)/3=0/3\]

Answer 8

\[x^{2}+y^{2}+14x+14y+62=0\]

Answer 9

ok now tell me what to do

Answer 10

\[(x+a)^2+(x+b)^2=c\] a,b,c\[a,b,c \in R\]

Answer 11

um whats a b c

Answer 12

for (x−5)^2+(y+2)^2=25

a=-5,b=2,c=25

Answer 13

um-337? idk this is hard

Answer 14

what is your a ,b and c values?

Answer 15

idk -5 2 25

Answer 16

\[r=\sqrt{c}\]

Answer 17

so is it 5
sooooooo is it 5

Answer 18

yes

Answer 19

can you solve the 3x^2+3y^2+42x+42y+186=0 on you own now ?Plz have a try