Question

If the circle \(x^2+y^2=a^2\) intersects the hyperbola \(xy=c^2\) in four points \((x_1,y_1)\) for \(i=1,2,3,4\), then \(y_1+y_2+y_3+y_4\) equals

• A. \(0\)
• B. \(c^2\)
• C. \(a\)
• D. \(c^4\)

Hint:

When intersection points are symmetric about origin, coordinates occur as \((x,y)\) and \((-x,-y)\), so their sums cancel.

Answer 1

The Correct Option is A

Step 1: Understand symmetry of curves.
Both curves \(x^2+y^2=a^2\) and \(xy=c^2\) are symmetric about both axes.
Step 2: Intersection points occur in symmetric pairs.
If \((x,y)\) is a solution, then because \(x^2+y^2\) unchanged and \(xy=c^2\) unchanged for \((-x,-y)\),
\[ (-x,-y) \] is also a solution.
Similarly, because \(xy=c^2\) requires \(x,y\) same sign, points lie in 1st and 3rd quadrants only, giving pairs:
\[ (x,y),\;(-x,-y) \]
Step 3: Add all y-coordinates.
Since y-values appear as \(y\) and \(-y\) in pairs, total sum is:
\[ y_1+y_2+y_3+y_4=0 \]
Final Answer:
\[ \boxed{0} \]