How do I get the length of A-B?

Have you considered using the Law of Cosines?
 
MarkFL said:
Have you considered using the Law of Cosines?
Click to expand...
Yes, cos a =( a^2 + b ^2 - c^2 )/2ab, I know that a=b = r, but how do I obtain the c? I can't get the value of cos a if I can't get the c, sorry for being rude, just want to clarify things up
 
I would write:

[MATH]\overline{AB}=r\sqrt{2(1-\cos(\alpha))}[/MATH]
 
I assumed, based on the thread title, that you wanted to know the chord length.

So, the equation of the circle is:

[MATH]x^2+y^2-4x-2y=31[/MATH]
Or, in standard form:

[MATH](x-2)^2+(y-1)^2=6^2[/MATH]
What is the equation of the line which passes through \(A\) and \(B\)?
 
MarkFL said:
I assumed, based on the thread title, that you wanted to know the chord length.

So, the equation of the circle is:

[MATH]x^2+y^2-4x-2y=31[/MATH]
Or, in standard form:

[MATH](x-2)^2+(y-1)^2=6^2[/MATH]
What is the equation of the line which passes through \(A\) and \(B\)?
Click to expand...
3x+4y+5=0
 
MethMath11 said:
3x+4y+5=0
Click to expand...

Okay, I would express this line in the form:

[MATH]y=-\frac{3x+5}{4}[/MATH]
Next, substitute for \(y\) in the equation of the circle:

[MATH](x-2)^2+\left(-\frac{3x+5}{4}-1\right)^2=6^2[/MATH]
Now, solve this resulting quadratic in \(x\) and use the equation of the line above to get the \((x,y)\) coordinates of the points of intersection between the line and the circle. What do you find?
 
MarkFL said:
Okay, I would express this line in the form:

[MATH]y=-\frac{3x+5}{4}[/MATH]
Next, substitute for \(y\) in the equation of the circle:

[MATH](x-2)^2+\left(-\frac{3x+5}{4}-1\right)^2=6^2[/MATH]
Now, solve this resulting quadratic in \(x\) and use the equation of the line above to get the \((x,y)\) coordinates of the points of intersection between the line and the circle. What do you find?
Click to expand...
So there's no quicker way than that. Thank you btw and now I have no idea how to Finnish it under 3/4 minutes
 
An alternative method would be to find the distance from point (2,1) to line 3x + 4y + 5 = 0 (there is a simple formula for that, or vector methods), and use simple trig to find angle q/2, then find its tangent. This works out very nicely.
 
Dr.Peterson said:
An alternative method would be to find the distance from point (2,1) to line 3x + 4y + 5 = 0 (there is a simple formula for that, or vector methods), and use simple trig to find angle q/2, then find its tangent. This works out very nicely.
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S = |3.2 +4.1+ 5| / √(3^2+ 4^2) ?
 
Well, at least carry out the calculation you showed. Clearly that's what comes next.

Then write that number in the picture, on the segment from P to the midpoint of AB. Call that midpoint C. What kind of triangle is ABC?

General principle: When you don't see the whole process, take one step, then another, and as you go, you have a better chance of seeing the way ahead.
 
MethMath11 said:
So there's no quicker way than that. Thank you btw and now I have no idea how to Finnish it under 3/4 minutes
Click to expand...

Solving that quadratic, you wind up with the points:

[MATH]\left(\frac{1\pm12\sqrt{3}}{5},\frac{-7\mp9\sqrt{3}}{5}\right)[/MATH]
Using the distance formula, we then find:

[MATH]\overline{AB}=6\sqrt{3}[/MATH]
Using the Law of Cosines, we next find:

[MATH]\cos(\alpha)=-\frac{1}{2}\implies \tan(\alpha)=-\sqrt{3}[/MATH]
 
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