Question

The abscissa of two points A and B are the roots of the equation x²+²ax−b² and their ordinates are the root of the equation x²+2px−q²=0. the equation of the circle with AB as diameter is

A. x²+y²+2ax+²py+b²+q²=0
B. x²+y²−2ax−²py−b²−q²=0
C. x²+y²+2ax+²py−b²−q²=0
D. None of these

Answer 1

The equation of the circle with diameter AB, whose endpoints have coordinates derived from the roots of the given quadratic equations, is found by determining its center using the averages of these roots and then constructing the circle equation. The correct answer is Option B: x²+y²−2ax−2py−b²−q²=0.

Step-by-step explanation:

The student is asked to find the equation of a circle with diameter AB, where A and B are points on the plane whose coordinates are the roots of two separate quadratic equations. The roots of x²+2ax-b²=0 are the x-coordinates (abscissas) of A and B, and the roots of x²+2px-q²=0 are the y-coordinates (ordinates) of A and B.

For a circle with diameter AB, the standard form of the equation is (x - h)² + (y - k)² = r², where (h, k) is the center of the circle and r is the radius. The midpoint of AB will be the center of the circle, which can be found using the averages of the x-coordinates and y-coordinates of A and B (the roots of the given quadratic equations).

Using the quadratic formula and the concept of the midpoint, we calculate the center (h, k) of the circle. After finding the center, we plug these values into the standard form and expand it to get the general form of the circle equation.

The correct option that represents the equation of the circle formed by diameter AB is Option B: x²+y²−2ax−2py−b²−q²=0.