Circle Chords: x^2 + y^2 - 12x + 6y - 20 = 0

[Monkeyseat]

Question: A circle with center C has equation x^2 + y^2 - 12x + 6y - 20 = 0

a) By using the method of completing the square, find the co-ordinates of C and the radius of the circle.

b) The origin O is the mid-point of a chord PQ of this circle.
i) Show that the gradient of the chord PQ is 2.
ii) Find the distance of O from the centre of the circle.
iii) Find the length of the chord PQ.

My working:

a) x^2 + y^2 - 12x + 6y - 20 = 0
(x-6)^2 - 36 + (y+3)^2 - 9 - 20 = 0
(x-6)^2 + (y+3)^2 = 65

Therefore, the centre is (6, -3) and the radius is sqrt. 65

I'm not sure how to do part b though, any suggestions?

Thanks.

 

[Deleted member 4993]

Re: Circle Chords

Question:

A circle with center C has equation x^2 + y^2 - 12x + 6y - 20 = 0

a) By using the method of completing the square, find the co-ordinates of C and the radius of the circle.

b) The origin O is the mid-point of a chord PQ of this circle.
i) Show that the gradient of the chord PQ is 2.
ii) Find the distance of O from the centre of the circle.
iii) Find the length of the chord PQ.

My working:

a)

x^2 + y^2 - 12x + 6y - 20 = 0
(x-6)^2 - 36 + (y+3)^2 - 9 - 20 = 0
(x-6)^2 + (y-3)^2 = 65

Therefore, the centre is (6, -4) and the radius is sqrt. 65

I'm not sure how to do B though, any suggestions?
Hint:

Perpendicular drawn from the center of the circle to a cord - bisects the cord

Thanks.

 

[Monkeyseat]

Re: Circle Chords

Question:

A circle with center C has equation x^2 + y^2 - 12x + 6y - 20 = 0

a) By using the method of completing the square, find the co-ordinates of C and the radius of the circle.

b) The origin O is the mid-point of a chord PQ of this circle.
i) Show that the gradient of the chord PQ is 2.
ii) Find the distance of O from the centre of the circle.
iii) Find the length of the chord PQ.

My working:

a)

x^2 + y^2 - 12x + 6y - 20 = 0
(x-6)^2 - 36 + (y+3)^2 - 9 - 20 = 0
(x-6)^2 + (y-3)^2 = 65

Therefore, the centre is (6, -4) and the radius is sqrt. 65

I'm not sure how to do B though, any suggestions?
Hint:

Perpendicular drawn from the center of the circle to a cord - bisects the cord

Thanks.

I know that but I'm still not sure what to do - I drew a diagram out and everything and didn't know where to start.

I'm guessing it's something to do with m1 * m2 = -1 but I only have 1 co-ordinate so I don't know how to find any gradients... :?

By the way, I meant (x-6)^2 + (y+3)^2 = 65 not (x-6)^2 + (y-3)^2 = 65 and (6, -3) not (6, -4).

 

[Deleted member 4993]

Re: Circle Chords

1) What is the middle-point of the cord in question?

2) If you connect this point to the center of the circle - what is the angle that this line and the given cord would make?

3) What is the slope of the line that you got in (2) - (by joining the middle point of the cord and the center of the circle) ?

4) What is the slope of the cord?

 

[Monkeyseat]

Re: Circle Chords

1) What is the middle-point of the cord in question?

2) If you connect this point to the center of the circle - what is the angle that this line and the given cord would make?

3) What is the slope of the line that you got in (2) - (by joining the middle point of the cord and the center of the circle) ?

4) What is the slope of the cord?

1) O

2) Right angle.

I don't know 3 and 4 - that's what I'm wondering - how do I find the gradients? I think they are perpendicular but that's it.

 

[Deleted member 4993]

Re: Circle Chords

1) What is the middle-point of the cord in question?

2) If you connect this point to the center of the circle - what is the angle that this line and the given cord would make?

3) What is the slope of the line that you got in (2) - (by joining the middle point of the cord and the center of the circle) ?

4) What is the slope of the cord?

1) (0,0)

2) Right angle.

I don't know 3 and 4 - that's what I'm wondering - how do I find the gradients? I think they are perpendicular but that's it.

The slope of the line joining

\(\displaystyle (x_1, y_1)\)

and

\(\displaystyle (x_2, y_2)\)

is

\(\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}\)

 

[Monkeyseat]

Re: Circle Chords

[quote="Subhotosh Khan":2xys1re2]1) What is the middle-point of the cord in question?

2) If you connect this point to the center of the circle - what is the angle that this line and the given cord would make?

3) What is the slope of the line that you got in (2) - (by joining the middle point of the cord and the center of the circle) ?

4) What is the slope of the cord?

1) (0,0)

2) Right angle.

I don't know 3 and 4 - that's what I'm wondering - how do I find the gradients? I think they are perpendicular but that's it.

The slope of the line joining

\(\displaystyle (x_1, y_1)\)

and

\(\displaystyle (x_2, y_2)\)

is

\(\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}\)[/quote:2xys1re2]

I get it now. I thought the "Origin O" was just the name of the mid-point lol, I didn't know it was actually (0,0). Thanks. I knew the formula, just didn't know the other co-ordinate.

Okay so I have done (i) and proved that it was 2 and I have also done (ii) and found that the distance from O to the centre is 3sqr. 5.

How do I do "iii) Find the length of the chord PQ"? I'm not sure again.

Thanks!

 

[Deleted member 4993]

Re: Circle Chords

Okay so I have done (i) and proved that it was 2 and I have also done (ii) and found that the distance from O to the centre is 3sqr. 5.

How do I do "iii) Find the length of the chord PQ"? I'm not sure again.

You know the radius - Now use Pythagorus....

Thanks!

 

[Monkeyseat]

Re: Circle Chords

Many thanks! I have done it. Using pythag I found OQ (end to mid-point) as 2sqrt. 5 so for PQ I doubled it and got 4sqrt. 5 - it looks right. Once again thanks.