Finding the Area of a Triangle Inside a Circle and Parallel Line: How to Solve?

[Kartik.]

1.There is a circle with the equation x^2 + y^2 - 2ax = 0. A line is drawn through the centre of the circle which is parallel to the line x+2y=0. and also intersects the circle at A and B. Find the area of the triangle AOB.

My attempt-
I calculated the slope of the given line(-1/2).So the equation of this line which cuts A and B on the circle would be y = -1/2x +c and as the line passes through the center of a circle which is drawn with origin as the center the equation turns y= -1/2x. The distance of A and B from the center is ax.So i would like to find the coordinates of A and B with that by-
ax = sqrt(x^2 + y^2) ---> which is true for both A and B. I'm stuck here.

 

[Mentallic]

Ok, there are quite a few things wrong there, so let's just start from the beginning.
You've calculated the slope of the line correctly, but you're misinterpreting what the circle should look like.

What is the general form of a circle? That is, if a circle has centre (h,k) with a radius of r, what equation describes this?

 

[Kartik.]

(x – h)^2 + (y – k)^2 = r^2
ok... the diameter would be sqrt(2ax) so the radius is (ax/sqrt(2)) ??

 

[Mentallic]

(x – h)^2 + (y – k)^2 = r^2

Yes!

ok... the diameter would be sqrt(2ax) so the radius is (ax/sqrt(2)) ??

No. You need to stop guessing! What we need to do is convert the given equation $$x^2 + y^2 - 2ax = 0$$ into the general form for the circle. Only once we have it in that form (exactly) can we then know what the centre and radius is equal to.

Now, how will we go about doing that? The general form has an (x-h)² term while we have an x² and 2ax term. Can you complete the square to convert x²+2ax into a perfect square?

 

[eumyang]

Yes!

No. You need to stop guessing! What we need to do is convert the given equation $$x^2 + y^2 - 2ax = 0$$ into the general form for the circle. Only once we have it in that form (exactly) can we then know what the centre and radius is equal to.

Now, how will we go about doing that? The general form has an (x-h)² term while we have an x² and 2ax term. Can you complete the square to convert x²+2ax into a perfect square?

Don't bother. The OP posted a duplicate thread https://www.physicsforums.com/showthread.php?t=628778, and DonAntonio provided most of the solution.

 

[Mentallic]

Don't bother. The OP posted a duplicate thread https://www.physicsforums.com/showthread.php?t=628778, and DonAntonio provided most of the solution.

The thread must've been deleted.

 

[HallsofIvy]

Yes, I deleted it since the appropriate place for this is here in homework.

Kartik, complete the square to determine where the center of the circle is, then you can write down the equation of the line and determine where it intersects the circle.

 

[Ray Vickson]

(x – h)^2 + (y – k)^2 = r^2
ok... the diameter would be sqrt(2ax) so the radius is (ax/sqrt(2)) ??

You are not thinking! You seem to be saying that the diameter of the circle changes as we move around the circle (to a new point (x,y) on the circle).

Do NOT just write down arbitrary formulas without understanding what they say; doing that will get you into trouble every time.

Rather than trying to write down an answer in some quick, shortcut way, you should approach a problem carefully, step-by-step. It takes longer, and may be boring, but is is much safer.

RGV

 

[Kartik.]

Thanks to everyone who tried to help and sorry for posting it two times :P.It was urgent back then. I have solved my question by applying the general form of a curve (ax^2+2hxy+by^2+2gx+2fy+c=0)
I just had my Circles intro class and applied whatever i did know :P (that x^2 + y^2 = (diameter)a^2((0,0)-centered), no matter in which form it is) :P

 

[Mentallic]

Thanks to everyone who tried to help and sorry for posting it two times :P.It was urgent back then. I have solved my question by applying the general form of a curve (ax^2+2hxy+by^2+2gx+2fy+c=0)

How exactly did you apply that formula to your problem?

I just had my Circles intro class and applied whatever i did know :P (that x^2 + y^2 = (diameter)a^2((0,0)-centered), no matter in which form it is) :P

I think it was pretty obvious what you were trying to do but clearly the equation you were given is different to the circle centred at (0,0) because a² is a positive constant value. Not a variable (no x or y involved).