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Question

Column IColumn II(A)If real numbers x and  y satisfy (x+5)2+(y12)2=81,then the minimum value of x2+y2 is(P)1(B)The line 3x+6y=k intersects the curve2x2+2xy+3y2=1 at point A and B.If the circle with AB as a diameter passesthrough the origin, then the value of  k2 is(Q)2(C)If two perpendicular tangents can bedrawn from the origin to the circlex26x+y22py+17=0 , thenthe value of |p| is(R)4(D)If the circlesx2+y2+(3+sinβ)x+(2cosα)y=0 andx2+y2+(2cosα)x+2cy=0 touch each other, then the maximum value of c  is(S)5(T)7(U)9
Which of the following is the only CORRECT combination?
  1. (A)(R), (B)(U)
  2. (A)(S), (B)(P)
  3. (A)(Q), (B)(U)
  4. (A)(R), (B)(Q)


Solution

The correct option is A (A)(R), (B)(U)
(A)
Equation of circle
(x+5)2+(y12)2=81 


Distance between the origin and the centre of the circle
=52+122=13
Minimum value of x2+y2 
=13r=139=4
(A)(R)

(B)
3x+6y=k
3x+6yk=1     (1)
Now, 2x2+2xy+3y21=0
To get the equation of lines OA and OB, we homogenize the curve using equation (1).
2x2+2xy+3y2(3x+6yk)2=0


As AB is the diameter of circle, so OA is perpendicular to OB.
Therefore, sum of coefficients of x2 and y2 is equal to zero.
29k2+336k2=0545k2=0
k2=9
(B)(U)
 

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