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samnatha
Group Title
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)
 one year ago
 one year ago
samnatha Group Title
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)
 one year ago
 one year ago

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phi Group TitleBest ResponseYou've already chosen the best response.1
what's your idea?
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
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 ?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
dw:1355075215700:dw
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
i have no idea what your talking about sorry you kind of lost me
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
I changed my mind. dw:1355075548150:dw
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
yes i get that but how do i get the equation ?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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= 3x6
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
could you go through that slower because we have been doing a different way than that ?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
maybe we should do it the way you are learning to do it. how far did you get?
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
well i subed in all my points and got the three equations which are 6g  4f + c =0 2f + c =0 2g + f = 0
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
3 equations and 3 unknowns is good. I would use "elimination" to solve them
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
i tried that and i got g = 2 but the answer in the back of the back of the book says its not that ?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
I think g should be 1 are you accounting for 2gx in front of x?
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
how did you get that ?
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
the answer in the back of the book says the equation is x^2 + y^2 + 2x + + 6y + 5 = 0
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
yes, the book is correct.
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
i din't get the radius or anything ?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
ok the first one for the point (3. 2) i did this 9 + 4 + 2g(3) + 2f(2) + c = 0
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
the second one (0, 1) i did 1 + 2g(0) + 2f(1) + c =0
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
and for the third i subed in (g, f) and i got this 2(g) y(f) + 4 = 0
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
for the 3rd equation I would would replace x with g and y with f to get \[g^2 + f^2 2g^22f^2+c=0 \]
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
that simplifies to \[ g^2 + f^2 c=0 \]
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
we never sub it into the formula usually
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
where do you sub in (g, f)? what do you sub this into?
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
the line 2x  y + 4 =0 to get the third equation
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
the line you want has slope 1/2 and goes thru point (3,2) y+2 = (1/2)(x+3)
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
oooh right so how would i do it then ?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
they give you 2x  y + 4 =0 in slopeintercept 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
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
how did you get the other slope ?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
the slope of lines that are perpendicular are "negative reciprocals" if one has slope m that other has slope 1/m
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
ok i see it now you flip the slope upside down and change the signs
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
so what would i do next ?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
now sub in (g,f) into the equation x+2y+7=0 now you have 3 eqs solve for g,f and c
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
where did you get x + 2y + 7 =0 from ?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
so do i do that with the other equations aswell ?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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+2f7=0 < 2f +g = 7
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
ah ok i get it now :) thanks
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
I got g=1, f=3, and c=5
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
ok thanks
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
can you solve the 3 eqs?
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
i'm just doing it now
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
i got 19/5 for g but i don't think thats right ?
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
wait it don't matter i see my mistake g is 1
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
It is a pain doing 3 eq and 3 unknowns
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
the i got 6 for f ? is that right ?
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
oh no wait i got it haha sorry i make really stupid mistakes
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
i got the answer thank you so much for the help :)
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.1
yupe i got all that thank you for your help would not have got it without you :)
 one year ago
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