finding equation of a circle and parabola

Jan 2010
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Finding the general and standard form equations for the circle and parabola passing through thee points (4,6), (-6,2), and (1,-3).

I used the y=Ax^2+Bx+C to find a,b,c.

(4,6)=> 6=A(4)^2+B(4)+C
(-6,2)=> 2=A(-6)^2+B(-6)+C
(1,-3)=> -3A(1)^2+B(1)+C

then,
a=13/15, b=8/7, and c=-158/35

so the equation for parabola:
\(\displaystyle y=(13/35)x^2+(8/7)x-(158/35)\)

how do I make this equation into this equation f(x)a(x-h)^2+k without the vertex?
 
Last edited:

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
Finding the general and standard form equations for the circle and parabola passing through thee points (4,6), (-6,2), and (1,-3).

I used the y=Ax^2+Bx+C to find a,b,c.

(4,6)=> 6=A(4)^2+B(4)+C
(-6,2)=> 2=A(-6)^2+B(-6)+C
(1,-3)=> -3A(1)^2+B(1)+C

then,
a=13/15, b=8/7, and c=-158/35

so the equation for parabola:
\(\displaystyle y=(13/35)x^2+(8/7)x-(158/35)\)

how do I make this equation into this equation f(x)a(x-h)^2+k without the vertex?
x-value of the vertex ... \(\displaystyle h = \frac{-b}{2a}\)

... and \(\displaystyle k = f(h)\)
 
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Jan 2010
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x-value of the vertex ... \(\displaystyle h = \frac{-b}{2a}\)

... and \(\displaystyle k = f(h)\)

I see, thanks!

Should I use the same a,b,c value from parabola for finding the equation for a circle?
 

skeeter

MHF Helper
Jun 2008
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North Texas
Should I use the same a,b,c value from parabola for finding the equation for a circle?
the equation for a circle is \(\displaystyle (x-h)^2 + (y-k)^2 = r^2
\) ... where will that idea get you?

I would find the circle's center (h,k) by locating the intersection of the perpendicular bisectors of at least two chords formed by the three points.
 
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Jan 2010
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the equation for a circle is \(\displaystyle (x-h)^2 + (y-k)^2 = r^2
\) ... where will that idea get you?

I would find the circle's center (h,k) by locating the intersection of the perpendicular bisectors of at least two chords formed by the three points.

I tried to find the equation for the circle and parabola from this three points (4,6),(-6,2), and (1,-3). this is what i get:

for parabola:
general form=
\(\displaystyle y=\frac{13}{35}x^2+\frac{8}{7}x-\frac{158}{35}\)
standard form=
\(\displaystyle y=\frac{13x^2-158}{35}+\frac{8x}{7}\)

Vertex: (158,8x/7)
focus: (58, 8x/7, -35/4)
directrix: y=8x/7 -35/4


for circle:

\(\displaystyle (x-\frac{77}{6})^2+(y-\frac{85}{22})^2 = 13.4^2 \)

radius = 13.4
center= (77/6, 85/22)


Am I right? please help me out... thanks!
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
I tried to find the equation for the circle and parabola from this three points (4,6),(-6,2), and (1,-3). this is what i get:

for circle:

\(\displaystyle (x-\frac{77}{6})^2+(y-\frac{85}{22})^2 = 13.4^2 \)

radius = 13.4
center= (77/6, 85/22)
sub in the values x = 4 y = 6 into the circle equation ... does it work?
 
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Jun 2009
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The general equation of the circle passing through (4, 6), (-6, 2) and (1, -3) is

\(\displaystyle x^2 + y^2 +2gx + 2fy + c = 0\)

Substitute the points in the above equation, and solve for g, f and c. I got the equation of the circle as

\(\displaystyle 13x^2 + 13y^2 + 10x -32y - 36 = 0\)

Check it.
Consider any one point as the common point for circle and parabola.
Let it be the vertex. Find the equation of the tangent to the circle at the vertex in the form ax +by + c = 0. Diretrix of the parabola will be parallel to this tangent in the form ax + by + c' = 0.
Let (h, k) be the focus. In the parabola, the distance between any point and the focus is equal the distance between the point and diretrix.

\(\displaystyle (h-4)^2 + (k-6)^2 = \frac{(4a + 6b + c')^2}{(a^2 + b^2)}\)

Write down three equations for three points and solve for h, k and c'
 
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Jan 2010
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Now we need to find the circle: (x - h)^2 + (y - k)^2 = r^2
center = (h,k)
radius = r

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So we solve another system of 3 equations:
(4 - h)^2 + (6 - k)^2 = r^2
(-6 - h)^2 + (2 - k)^2 = r^2
(1 - h)^2 + (-3 - k)^2 = r^2

[(4 - h)^2 + (6 - k)^2 = r^2] - [(-6 - h)^2 + (2 - k)^2 = r^2] = (-20h - 8k + 12 = 0)
[(4 - h)^2 + (6 - k)^2 = r^2] - [(1 - h)^2 + (-3 - k)^2 = r^2] = (-6h - 18k + 42 = 0)

Now solve the system of two equations:
(-20h - 8k + 12 = 0)
(-6h - 18k + 42 = 0)

h = (-5/13)
k = (32/13)

r = sqrt(5365)/13

x + (5/13)]^2 + [y - (32/13)]^2 = [sqrt(5365)/13]^2

this is what i got.......