Question

Let the abscissae of the two points P and Q be the roots of 2x² – rx + p = 0 and the ordinates of P and Q be the roots of x² – sx – q = 0. If the equation of the circle described on PQ as diameter is 2(x² + y²) – 11x – 14y – 22 = 0, then 2r + s – 2q + p is equal to

Comments

Hag diya behnchod explain toh kar bsdk kaise aaya . Chutiya app

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general eqn of circle (diametric form)

(x-x1) (x-x2) +(y-y1)(y-y2)=0          

simplify it
x^2 + y^2  -x(x1+x2) -y(y1+y2) +(x1x2 + y1y2)=0           (eqn 1)


 devide the whole eqn given in ques by 2 ...then the given eqn become
 
x^2 + y^2  -11/2 x  -7y  -11 =0             (eqn2)
 
now compare eqn 1 & eqn 2

x1+x2 = 11/2    (eqn3)
y1+y2= 7               (eqn4)
x1x2 + y1y2 = -11                (eqn5)


as it is given in ques the root of two quadratic eqn are the abscissae and ordinate of point P&Q
so,
x1 + x2 = r/2       (eqn6)
x1x2= -p/2           (eqn7)


y1+y2 = s            (eqn8)
y1y2 = -q            (eqn9)

(by basic concept of quadratic eqn)

solve eqn 3,4 ,5 with 6,7,8,9
you will get the value of    r , s , -2q+p
final answer will be 7

hope you get it .

Answer 1

Equation of the circle with PQ as diameter is 2(x² + y²) – rx – 2sy + p – 2q = 0 on comparing with the given equation

r = 11, s = 7 

p – 2q = – 22

\(\therefore\) 2r + s - 2q + p

 = 22 + 7 - 22

 = 7