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)

- anonymous

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- phi

what's your idea?

- 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

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- anonymous

i have no idea what your talking about sorry you kind of lost me

- phi

I changed my mind.
|dw:1355075548150:dw|

- anonymous

yes i get that but how do i get the equation ?

- 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

could you go through that slower because we have been doing a different way than that ?

- phi

maybe we should do it the way you are learning to do it. how far did you get?

- 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

3 equations and 3 unknowns is good.
I would use "elimination" to solve them

- 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

I think g should be 1
are you accounting for 2gx in front of x?

- anonymous

how did you get that ?

- anonymous

the answer in the back of the book says the equation is
x^2 + y^2 + 2x + + 6y + 5 = 0

- phi

yes, the book is correct.

- anonymous

i din't get the radius or anything ?

- 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

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

ok the first one for the point (-3. -2) i did this
9 + 4 + 2g(-3) + 2f(-2) + c = 0

- anonymous

the second one (0, -1) i did
1 + 2g(0) + 2f(-1) + c =0

- anonymous

and for the third i subed in (-g, -f)
and i got this
2(-g) -y(-f) + 4 = 0

- 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

that simplifies to
\[ g^2 + f^2 -c=0 \]

- anonymous

we never sub it into the formula usually

- phi

where do you sub in (-g, -f)? what do you sub this into?

- anonymous

the line 2x - y + 4 =0 to get the third equation

- 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

the line you want has slope -1/2 and goes thru point (-3,-2)
y+2 = (-1/2)(x+3)

- anonymous

oooh right so how would i do it then ?

- 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

how did you get the other slope ?

- phi

the slope of lines that are perpendicular are "negative reciprocals" if one has slope m
that other has slope -1/m

- anonymous

ok i see it now you flip the slope upside down and change the signs

- anonymous

so what would i do next ?

- 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

where did you get x + 2y + 7 =0 from ?

- 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

so do i do that with the other equations aswell ?

- 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

ah ok i get it now :) thanks

- phi

I got g=1, f=3, and c=5

- 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

ok thanks

- phi

can you solve the 3 eqs?

- anonymous

i'm just doing it now

- anonymous

i got 19/5 for g but i don't think thats right ?

- anonymous

wait it don't matter i see my mistake g is 1

- phi

It is a pain doing 3 eq and 3 unknowns

- anonymous

the i got 6 for f ? is that right ?

- anonymous

oh no wait i got it haha sorry i make really stupid mistakes

- anonymous

i got the answer thank you so much for the help :)

- 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

yupe i got all that thank you for your help would not have got it without you :)

- phi

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