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 2 years ago
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)
 2 years ago
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)

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samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1am 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 ?

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1i have no idea what your talking about sorry you kind of lost me

phi
 2 years ago
Best ResponseYou've already chosen the best response.1I changed my mind. dw:1355075548150:dw

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1yes i get that but how do i get the equation ?

phi
 2 years ago
Best ResponseYou've already chosen the best response.1the 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

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

phi
 2 years ago
Best ResponseYou've already chosen the best response.1maybe we should do it the way you are learning to do it. how far did you get?

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

phi
 2 years ago
Best ResponseYou've already chosen the best response.13 equations and 3 unknowns is good. I would use "elimination" to solve them

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

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

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1how did you get that ?

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

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1i din't get the radius or anything ?

phi
 2 years ago
Best ResponseYou've already chosen the best response.1my 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
 2 years ago
Best ResponseYou've already chosen the best response.1in 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?

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

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

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

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

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

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1we never sub it into the formula usually

phi
 2 years ago
Best ResponseYou've already chosen the best response.1where do you sub in (g, f)? what do you sub this into?

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

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

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1oooh right so how would i do it then ?

phi
 2 years ago
Best ResponseYou've already chosen the best response.1they 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

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1how did you get the other slope ?

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

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1ok i see it now you flip the slope upside down and change the signs

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1so what would i do next ?

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

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

phi
 2 years ago
Best ResponseYou've already chosen the best response.1After 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

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1so do i do that with the other equations aswell ?

phi
 2 years ago
Best ResponseYou've already chosen the best response.1your 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

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1ah ok i get it now :) thanks

phi
 2 years ago
Best ResponseYou've already chosen the best response.1I 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.

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1i got 19/5 for g but i don't think thats right ?

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1wait it don't matter i see my mistake g is 1

phi
 2 years ago
Best ResponseYou've already chosen the best response.1It is a pain doing 3 eq and 3 unknowns

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1the i got 6 for f ? is that right ?

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1oh no wait i got it haha sorry i make really stupid mistakes

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.1i got the answer thank you so much for the help :)

phi
 2 years ago
Best ResponseYou've already chosen the best response.1no. 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

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