How Do You Find the Equation of a Line Through a Point and a Circle's Center?

[mathdad]

Find an equation of the line that passes through (3, -5) and through the center of the circle 4x^2 + 8x + 4y^2 - 24y + 15 = 0.

1. I got to find the center of the circle.

2. I got to find the distance between the center of the circle and the given point. The distance is the radius.

3. I must then write the equation in standard form.

Yes?

 

[MarkFL]

Find an equation of the line that passes through (3, -5) and through the center of the circle 4x^2 + 8x + 4y^2 - 24y + 15 = 0.

1. I got to find the center of the circle.

Yes, that will given you two points on the line, allowing you to determine the equation of the line.

2. I got to find the distance between the center of the circle and the given point. The distance is the radius.

No, you don't need to know the radius of the circle, although you will likely have it when finding the center of the circle.

3. I must then write the equation in standard form.

Yes?

The problem statement simply says to find an equation of the line...the form in which you give it is up to you. :)

 

[HallsofIvy]

Find an equation of the line that passes through (3, -5) and through the center of the circle 4x^2 + 8x + 4y^2 - 24y + 15 = 0.

1. I got to find the center of the circle.

4(x^2+ 2x)+ 4(y^2- 6y)= -15. Complete the two squares.

2. I got to find the distance between the center of the circle and the given point. The distance is the radius.

No, you don't need to do find the distance. Further, since 4(9)+ 8(3)+ 5(25)- 24(-5)+ 15 is NOT 0, the given point does NOT lie on the circle. The distance between the two points is NOT the radius of the circle.

3. I must then write the equation in standard form.

Yes?

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Are you required to write it in "standard form"? Any (non-vertical) line can be written in the form y= ax+ b. Taking the two points, the center of the circle and (3, -5), as (x, y) gives two equations to solve for a and b.

 

[mathdad]

Yes, that will given you two points on the line, allowing you to determine the equation of the line.
No, you don't need to know the radius of the circle, although you will likely have it when finding the center of the circle.
The problem statement simply says to find an equation of the line...the form in which you give it is up to you. :)

Steps:

1. Find center of the circle.

2. Find the slope m having the given point and center of circle.

3. Use the point-slope formula using the slope found in step 3 and one of the points.

4. Isolate y to establish the equation of the line.

 

[mathdad]

4x^2 + 8x + 4y^2 - 24y + 15 = 0

x^2 + 2x + 1 + y^2 - 6y + 9 = - 15/4 + 1 + 9

(x + 1)(x + 1) + (y - 3)(y - 3) = 25/4

(x + 1)^2 + (y - 3)^2 = 25/4

Center: (-1, 3)

Let m = slope

m = (-5 -3)/(4)

m = -8/4

m = -2

I will use the given point (3, -5).

y -(-5) = -2(x - 3)

y + 5 = -2x + 6

y = -2x + 6 - 5

y = -2x + 1

Correct?

 

[MarkFL]

Looks good to me:

View attachment 7261

 

[mathdad]

Looks good to me:

Thanks. I am planning to review geometry in addition to precalculus. I will start posting geometry questions beginning next week. I would like to see more geometric pictures like the one posted here.