2. If the abscissae of points A, B are the roots of
the equation x2 + 2ax - b2 = 0 and ordinates
of A, B are roots of y2 + 2py-q2 = (), then find
the equation of the circle for which AB is a
diameter
(MARCH-2014)
Answers
Step-by-step explanation:
The abscissa of two points A and B are the roots of the equation x^+2ax-b^=0 and their ordinates are the roots of the equation y^+2py-q^=0, then what is the radius of the circle with AB as a diameter?
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4 Answers

Ved Prakash Sharma, former Lecturer at Sbm Inter College, Rishikesh (1971-2007)
Answered December 30, 2018
Let coordinates of A and B are (x1,y1) and (x2,y2).
x1 and x2 are the roots of x^2 +2ax -b^2=0, therefore
x1+x2= -2a……………(1). and. x1.x2= -b^2……………..(2)
y1 ,y2 are the roots of y^2+2py-q^2=0 , therefore
y1+y2=-2p……………..(3). and. y1.y2=-q^2……………..(4)
Length of diameter AB =√[(x1-x2)^2+(y1-y2)^2]
Formula used. [(a-b)^2=(a+b)^2–4a.b]
Diameter AB =√[(x1+x2)^2–4x1.x2+(y1+y2)^2–4y1.y2]
= √[4a^2+4b^2+4p^2+4q^2]
= 2.√(a^2+b^2+p^2+q^2) units.
Radius = diameter/2 = 2.√(a^2+b^2+p^2+q^2)/2
Radius =√(a^2+b^2+p^2+q^2). units.
HOPE THIS WILL HELP YOU!!!!!!!!!!