Multiple choice

The abcissae of two points A and B are the roots of the equation $\displaystyle x^{2}+2ax-b^{2}=0 $ and their ordinates are the roots of the equation $\displaystyle x^{2}+2px-q^{2}=0 $ The radius of the circle with AB as diameter is

  1. $\displaystyle \sqrt{a^{2}+b^{2}+p^{2}+q^{2}} $
  2. $\displaystyle \sqrt{a^{2}+p^{2}} $
  3. $\displaystyle \sqrt{b^{2}+q^{2}} $
  4. $\displaystyle \sqrt{a^{2}-b^{2}+p^{2}-q^{2}} $
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A Correct answer
Explanation

The difference between the two abscissae is 2sqrt(a^2 + b^2), and the difference between the ordinates is 2sqrt(p^2 + q^2). The diameter squared is therefore 4(a^2 + b^2 + p^2 + q^2), so the radius is sqrt(a^2 + b^2 + p^2 + q^2).

AI explanation

Let the roots of the first equation be x1 and x2, so their sum is -2a and product is -b^2. Let the roots of the second equation be y1 and y2, so their sum is -2p and product is -q^2. The coordinates of the endpoints of the diameter are (x1, y1) and (x2, y2). The square of the diameter AB^2 is (x1 - x2)^2 + (y1 - y2)^2. We know (x1 - x2)^2 = (x1 + x2)^2 - 4*x1*x2 = (-2a)^2 - 4*(-b^2) = 4(a^2 + b^2). Similarly, (y1 - y2)^2 = (-2p)^2 - 4*(-q^2) = 4(p^2 + q^2). Thus, AB^2 = 4(a^2 + b^2 + p^2 + q^2). Since the radius is half of the diameter, the radius is sqrt(a^2 + b^2 + p^2 + q^2).

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