Circles Chords and parallel lines

Question

I was going thru this question - Two Chords Ab, Cd of Lengths 5 Cm, 11 Cm Respectively of a Circle Are Parallel. If the Distance Between Ab and Cd is 3 Cm, Find the Radius of the Circle.

As per explanation provided in here I disagree - how can one be sure that cords are on one side of the semi-circle. I was trying the same method but judged that the chords are in each other hemisphere.

That means that the radius can vary based on distance between the chords if they are (i) on the same side of the center. (ii) on the opposite sides of the center.

In fact we get wrong answer if they are in opposite center - so how can we know from an approach when to consider chords on same / other side.

Answer 1

I would say it is not immediately clear that the two chords should be on the same side of the center. So you are right to prove that indeed it is impossible that they are on opposite sides of the center.

One way to prove this is to note that the radius $r$ of the circle satisfies $r\geq\tfrac{11}{2}$, because the circle has a chord of length $11$. Then by Pythagoras the distance from the chord of length $5$ to the center is $$\sqrt{r^2-\left(\tfrac52\right)^2}\geq\sqrt{\frac{11^2-5^2}{2^2}}=2\sqrt{6}>3,$$ so if the distance to the other chord is $3$ then it cannot be on the other side of the center.

Answer 2

Interesting question. Personally, I would make two sets of calculations: one for same semi-circle and another for opposing semi-circle. In that manner, potential impossibilities can be eliminated. Also, there might be scenarios where two correct solutions are revealed. I feel that the example failed to address this, so you were right to be astute to it. In fact, come to think of it, there will always be a solution in the same semi-circle. Then the question remains, what else can we do to determine whether there is an opposing semi-circle solution too?

Answer 3

Here's another approach. I believe the principle is correct to answer your question. The given example used lengths, but if displacements were used instead, then positive and negative values would come into play. I will clarify what I mean. With a circle centre O at the origin of an x-y grid, the equation is x^2 + y^2 = r^2. If we stick to horizontal chords, we can say that half a chord length is equal to the x-coordinate of where the chord intersects the circle (considering only x > 0). We could also agree to set up the smaller chord to lie above the x-axis, namely where y>0. We will call the smaller chord, "CHORD 2". Immediately we can set up 2 equations: A) x1^2 + y1^2 = r^2 B) x2^2 + y2^2 = r^2 We also know that y2-y1 = 3 So our equations become: A) x1^2 + y1^2 = r^2 B) x2^2 + (y1 + 3)^2 = r^2 Note that Equation A = Equation B = r^2 So, x1^2 + y1^2 = x2^2 + (y1 + 3)^2 Hence, for CHORD 1, (y1 + 3)^2 - y1^2 + (x2^2 - x1^2) = 0 Simplified, 6y1 = x1^2 - x2^2 - 9 Using x1 = 11/2 and x2 =5/2, we get: 6y1 = 121/4 - 25/4 - 9 = 15 So y1 = 15/6 = 5/2 > 0, so the solution is in the same semicircle. Now it is an easy matter to substitute x1= 11/2 and y1 = 5/2 into x^2 + y^2 = r^2 121/4 + 25/4 = r^2 So r = SQRT(36.5)

As for a generic formal to decide if the 2 chords lie in the same semicircle, Use 6y1 = x1^2 - x2^2 - d^2 where x1 = larger chord/2 , x2 = shorter chord/2 , d = perpendicular distance between the chords. If y1 > 0, then chords lie in same semi-circle. If y1 < 0, then chords lie in opposite semi-semicircles. I hope this helps.