Question

can i have some help please, will favourite fan and testimonial, you name it!

Answer 1

it's question 5

Answer 2

@ganeshie8 pleaseee

Answer 3

what do you know about radius of the circle that just touches the x axis ?

Answer 4

radius equals "y" coordinate of the center right ?

Answer 5

these are my attempts @ganeshie8 but i have a problem at the end there

Answer 6

paer i looks good!

Answer 7

what about paper 2? where am i going wrong?

Answer 8

sry i meant to say "part i" looks good

Answer 9

oh what about part 2??

Answer 10

its wrong

Answer 11

looks you have interpreted the problem incorrectly

Answer 12

\[\sf x^2 + 2gx +y^2 +2fy +c=0\]\[\sf (x^2 +2gx+g^2)+(y^2 +2fy +f^2)+c=0\]\[\sf (x+g)^2 +(y+f)^2 + -g^2-f^2+c=0\] Something like that for part i using completing the squares

Answer 13

oh, how should i have interpreted it?

Answer 14

ok thnks jhannybean

Answer 15

you're assuming the given points A and B are centers which is wrong

Answer 16

i didn't even know i was doing that? how was i even doing that, i just thought i had to find equations to find out the value of the 3 variables g,f, c ??

Answer 17

R is not |-4| or |-2|

Answer 18

you need to solve below system :
\[3^2+4^2+2g(3)+2f(4) +c = 0 \tag{1}\]
\[5^2+2^2+2g(5)+2f(2) +c = 0 \tag{2}\]
\[g^2 = c \tag{3}\]

Answer 19

solving them gives you two sets of solutions for g,f,c :
http://www.wolframalpha.com/input/?i=solve++++3%5E2%2B4%5E2%2B2g*%283%29%2B2f*%284%29+%2Bc+%3D+0%2C+5%5E2%2B2%5E2%2B2g*%285%29%2B2f*%282%29+%2Bc+%3D+0%2Cg%5E2+%3D+c

Answer 20

so do you just sub g^2 into your first 2 equations?

Answer 21

yes just replace "c" by g^2 in first two equations and solve g, f first

Answer 22

wait, i sub g^2 into both equations then take the second equation from the 1st one?

Answer 23

hey no wait, thats really a bad idea

Answer 24

oh so what should i do then ?

Answer 25

i can't think of any other easy way, go ahead... do it :)

Answer 26

yeah i got -4g+4f=0 when i solved them, like earlier on... now what do i do? :)

Answer 27

If you're still interested in the completing the square method.. continue from here :) \[\sf (x+g)^2 +(y+f)^2 + -g^2-f^2+c=0\] You've gotten to complete the square but now you're stil wondering, "how do g, f , and c relate to a circle?" We know \(\sf (x+h)^2 +(y+k)^2 =r^2\)- equation of a circle.

Answer 28

But in our question, we have \(\sf (x+g)^2 +(y+f)^2 \color{red}{-g^2-f^2+c}=0\)

Relating the equation of the circle to our equation, we have:\[\color{blue}{(x+h)^2} +\color{green}{(y+k)^2} =\color{purple}{r^2}\]\[\color{blue}{(x+g)^2}+\color{green}{(y+f)^2}= \color{purple}{g^2+f^2-c}\]

Answer 29

i'm so confused ....

Answer 30

@ganeshie8 what do i do after taking them away?

Answer 31

plugging c = g^2 in first two equations gives u :

25 + 6g + 8f + g^2 = 0
29 + 10g + 4f + g^2 = 0

subtracting both gives you :

4 + 4g - 4f = 0

g = f - 1