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nycguitarguy

Member

Registered: 2024-02-24

Posts: 552

Find Equation of the Line

Find an equation of the line containing the centers of the two circles below.

TWO CIRCLES

x^2 + y^2 - 4x + 6y + 4 = 0

AND

x^2 + y^2 + 6x + 4y + 9 = 0

Can someone get me started here?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,199

Re: Find Equation of the Line

Get the circle equations in the form:

(x+g)^2 + (y+h)^2 = r^2

(g,h) is the centre [and (g' h') for the other.]

So you can create the equation of the line through those two points.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

nycguitarguy

Member

Registered: 2024-02-24

Posts: 552

Re: Find Equation of the Line

Get the circle equations in the form:

(x+g)^2 + (y+h)^2 = r^2

(g,h) is the centre [and (g' h') for the other.]

So you can create the equation of the line through those two points.

Bob

Ok. Will do.

amnkb

Member

Registered: 2023-09-19

Posts: 253

Re: Find Equation of the Line

Last time you said you wanted steps but not answers so Bob taught me to hide the steps
so Ill do that again I hope thats ok w/ you

Find an equation of the line containing the centers of the two circles below.

x^2 + y^2 - 4x + 6y + 4 = 0

x^2 + y^2 + 6x + 4y + 9 = 0

Can someone get me started here?

nycguitarguy

Member

Registered: 2024-02-24

Posts: 552

Re: Find Equation of the Line

Last time you said you wanted steps but not answers so Bob taught me to hide the steps
so Ill do that again I hope thats ok w/ you

Find an equation of the line containing the centers of the two circles below.

x^2 + y^2 - 4x + 6y + 4 = 0

x^2 + y^2 + 6x + 4y + 9 = 0

Can someone get me started here?

Very good. Thanks.

KerimF

Member

From: Aleppo-Syria

Registered: 2018-08-10

Posts: 167

Re: Find Equation of the Line

May I add.

The equation of a straight line could be written as:
y = m*x + c
where
m is the slope of the line
c is where the line crosses the Y axis (usually the vertical one), that is when x=0.

In case two points on it are known, as (x1, y1) and (x2, y2), its slope 'm' could be calculated as:
m = (y2 - y1) / (x2 - x1)
This gives us:
y = (y2 - y1) / (x2 - x1)*x + c

To find c, we know that the coordinates of each point satisfies the line's equation. So, we may write:
y1 = (y2 - y1) / (x2 - x1)*x1 + c
or
y2 = (y2 - y1) / (x2 - x1)*x2 + c

Let us find c:
c = y1 - (y2 - y1) / (x2 - x1)*x1
c = [y1*(x2 - x1) - (y2 - y1)*x1] / (x2 - x1)
c = (y1*x2 - y1*x1 - y2*x1 + y1*x1) / (x2 - x1)
c = (y1*x2 - y2*x1) / (x2 - x1)
or
c = y2 - (y2 - y1) / (x2 - x1)*x2
c = [y2*(x2 - x1) - (y2 - y1)*x2] / (x2-x1)
c = (y2*x2 - y2*x1 - y2*x2 + y1*x2) / (x2-x1)
c = (- y2*x1 + y1*x2) / (x2-x1)
c = (y1*x2 - y2*x1) / (x2-x1)

We get the same value of c, by using the first or second point.

Therefore, y = m*x + c could be written by using the coordinates of the line's: two points as:
y = x *(y2 - y1) / (x2 - x1) + (y1*x2 - y2*x1) / (x2-x1)

y = [x *(y2 - y1) + (y1*x2 - y2*x1)] / (x2-x1)

In practice, this last form is not used directly.
The numerical value of the slope is found first.
Then the value of m and the (x, y) of one of the two points are used in y=mx+c to find the value of c.
For example, as amnkb calculated m, m = -1/5. Then, by replacing y and x with the coordinates of the first point (2, -3), we get:
y = -x/5 + c
-3 = -2/5 + c
c = -3 + 2/5
c = (-15 + 2)/5
c = -13/5
Therefore, the equation of the line is (as found by amnkb):
y = -x/5 - 13/5

Last edited by KerimF (2024-03-05 18:57:19)

nycguitarguy

Member

Registered: 2024-02-24

Posts: 552

Re: Find Equation of the Line

May I add.

The equation of a straight line could be written as:
y = m*x + c
where
m is the slope of the line
c is where the line crosses the Y axis (usually the vertical one), that is when x=0.

In case two points on it are known, as (x1, y1) and (x2, y2), its slope 'm' could be calculated as:
m = (y2 - y1) / (x2 - x1)
This gives us:
y = (y2 - y1) / (x2 - x1)*x + c

To find c, we know that the coordinates of each point satisfies the line's equation. So, we may write:
y1 = (y2 - y1) / (x2 - x1)*x1 + c
or
y2 = (y2 - y1) / (x2 - x1)*x2 + c

Let us find c:
c = y1 - (y2 - y1) / (x2 - x1)*x1
c = [y1*(x2 - x1) - (y2 - y1)*x1] / (x2 - x1)
c = (y1*x2 - y1*x1 - y2*x1 + y1*x1) / (x2 - x1)
c = (y1*x2 - y2*x1) / (x2 - x1)
or
c = y2 - (y2 - y1) / (x2 - x1)*x2
c = [y2*(x2 - x1) - (y2 - y1)*x2] / (x2-x1)
c = (y2*x2 - y2*x1 - y2*x2 + y1*x2) / (x2-x1)
c = (- y2*x1 + y1*x2) / (x2-x1)
c = (y1*x2 - y2*x1) / (x2-x1)

We get the same value of c, by using the first or second point.

Therefore, y = m*x + c could be written by using the coordinates of the line's: two points as:
y = x *(y2 - y1) / (x2 - x1) + (y1*x2 - y2*x1) / (x2-x1)

y = [x *(y2 - y1) + (y1*x2 - y2*x1)] / (x2-x1)

In practice, this last form is not used directly.
The numerical value of the slope is found first.
Then the value of m and the (x, y) of one of the two points are used in y=mx+c to find the value of c.
For example, as amnkb calculated m, m = -1/5. Then, by replacing y and x with the coordinates of the first point (2, -3), we get:
y = -x/5 + c
-3 = -2/5 + c
c = -3 + 2/5
c = (-15 + 2)/5
c = -13/5
Therefore, the equation of the line is (as found by amnkb):
y = -x/5 - 13/5

Great study notes. Thanks.