anonymous
  • anonymous
please help i've some idea of how to do it but i'm not sure ? Find the equation of the circle which passes through the points P(-3,-2) and Q(0, -1) and has the line 2x - y + 4 =0 as a tangent at the point P(-3,-2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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phi
  • phi
what's your idea?
anonymous
  • anonymous
am well you use the x^2 + y^2 + 2gx + 2fy _ c =0 formula and sub the points into to get the two equations and then to get the third one you sub in -g for x and -f for y ?
phi
  • phi
|dw:1355075215700:dw|

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anonymous
  • anonymous
i have no idea what your talking about sorry you kind of lost me
phi
  • phi
I changed my mind. |dw:1355075548150:dw|
anonymous
  • anonymous
yes i get that but how do i get the equation ?
phi
  • phi
the line normal to y= 2x+4 has slope -1/2 and goes thru pt (-3,-2) the segment that goes through pts (-3,-2) and (0,-1) has slope (-1+2)/(0+3)= 1/3 the normal will have slope -3 (-1/m) it goes thru the midpt (-3/2, -3/2), so its equation is y= -3x-6
anonymous
  • anonymous
could you go through that slower because we have been doing a different way than that ?
phi
  • phi
maybe we should do it the way you are learning to do it. how far did you get?
anonymous
  • anonymous
well i subed in all my points and got the three equations which are -6g - 4f + c =0 -2f + c =0 -2g + f = 0
phi
  • phi
3 equations and 3 unknowns is good. I would use "elimination" to solve them
anonymous
  • anonymous
i tried that and i got g = 2 but the answer in the back of the back of the book says its not that ?
phi
  • phi
I think g should be 1 are you accounting for 2gx in front of x?
anonymous
  • anonymous
how did you get that ?
anonymous
  • anonymous
the answer in the back of the book says the equation is x^2 + y^2 + 2x + + 6y + 5 = 0
phi
  • phi
yes, the book is correct.
anonymous
  • anonymous
i din't get the radius or anything ?
phi
  • phi
my way worked, but it is not the way to learn how to do your approach. I got (-1,-3) as the center and sqrt(5) for the radius
phi
  • phi
in your approach you find g, f, and c and get the equation you posted x^2 + y^2 + 2x + 6y + 5 = 0 we could (if we had to) figure out the center of the circle and the radius by completing the square for x^2 and y^2. But first, how did you set up your equations?
anonymous
  • anonymous
ok the first one for the point (-3. -2) i did this 9 + 4 + 2g(-3) + 2f(-2) + c = 0
anonymous
  • anonymous
the second one (0, -1) i did 1 + 2g(0) + 2f(-1) + c =0
anonymous
  • anonymous
and for the third i subed in (-g, -f) and i got this 2(-g) -y(-f) + 4 = 0
phi
  • phi
for the 3rd equation I would would replace x with -g and y with -f to get \[g^2 + f^2 -2g^2-2f^2+c=0 \]
phi
  • phi
that simplifies to \[ g^2 + f^2 -c=0 \]
anonymous
  • anonymous
we never sub it into the formula usually
phi
  • phi
where do you sub in (-g, -f)? what do you sub this into?
anonymous
  • anonymous
the line 2x - y + 4 =0 to get the third equation
phi
  • phi
that only makes sense if you use the line perpendicular to the tangent line (-g,-f) is the center of the circle. |dw:1355078200487:dw|
phi
  • phi
the line you want has slope -1/2 and goes thru point (-3,-2) y+2 = (-1/2)(x+3)
anonymous
  • anonymous
oooh right so how would i do it then ?
phi
  • phi
they give you 2x - y + 4 =0 in slope-intercept form that is y= 2x+4 (just add y to both sides) its slope is 2. the slope of the line we want is -1/m or in this case -1/2
anonymous
  • anonymous
how did you get the other slope ?
phi
  • phi
the slope of lines that are perpendicular are "negative reciprocals" if one has slope m that other has slope -1/m
anonymous
  • anonymous
ok i see it now you flip the slope upside down and change the signs
anonymous
  • anonymous
so what would i do next ?
phi
  • phi
now sub in (-g,-f) into the equation x+2y+7=0 now you have 3 eqs solve for g,f and c
anonymous
  • anonymous
where did you get x + 2y + 7 =0 from ?
phi
  • phi
After we figured out its slope was -1/2 we using the point slope formula or just write down y= mx+b and replace m with -1/2 , and (-3,-2) for (x,y) (the line goes thru this point) solve for b. then write the equation in standard form. -2 = (-1/2) * (-3) +b b= -2 - 3/2 = -7/2 y= (-1/2) x -7/2 multiply both sides by 2 2y = -x -7 add +x+7 to both sides x+2y+7 =0
anonymous
  • anonymous
so do i do that with the other equations aswell ?
phi
  • phi
your eqs are 9 + 4 + 2g(-3) + 2f(-2) + c = 0 <-- -4f -6g +c = -13 1 + 2g(0) + 2f(-1) + c =0 <-- -2f +c = -1 g+2f-7=0 <-------------------- 2f +g = 7
anonymous
  • anonymous
ah ok i get it now :) thanks
phi
  • phi
I got g=1, f=3, and c=5
phi
  • phi
I noticed your original equations were all set = to 0 (and they should not have been) so that is one other thing to watch out for.
anonymous
  • anonymous
ok thanks
phi
  • phi
can you solve the 3 eqs?
anonymous
  • anonymous
i'm just doing it now
anonymous
  • anonymous
i got 19/5 for g but i don't think thats right ?
anonymous
  • anonymous
wait it don't matter i see my mistake g is 1
phi
  • phi
It is a pain doing 3 eq and 3 unknowns
anonymous
  • anonymous
the i got 6 for f ? is that right ?
anonymous
  • anonymous
oh no wait i got it haha sorry i make really stupid mistakes
anonymous
  • anonymous
i got the answer thank you so much for the help :)
phi
  • phi
no. Here is how I do it. arrange alphabetically (and it is good the coeff of c is 1) c f g constant 1 -4 -6 -13 <-- -4f -6g +c = -13 1 -2 0 -1 <---- -2f +c = -1 0 2 1 7 <------ 2f +g = 7 subtract replace row 2 with row 2 - row 1 1 -4 -6 -13 0 2 6 12 0 2 1 7 now replace 3rd row with 3rd row - 2nd row 1 -4 -6 -13 0 2 6 12 0 0 -5 -5 now back solve -5g= -5 so g= 1 2f +6(1)= 12 2f =6 f=3 c -4(3) -6(1)= -13 c -18 = -13 c= 5
anonymous
  • anonymous
yupe i got all that thank you for your help would not have got it without you :)
phi
  • phi
yw

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