  Question

# Column IColumn II(A)If real numbers x and  y satisfy (x+5)2+(y−12)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 circlex2−6x+y2−2py+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?(A)→(R), (B)→(U)(A)→(S), (B)→(P)(A)→(Q), (B)→(U)(A)→(R), (B)→(Q)

Solution

## The correct option is A (A)→(R), (B)→(U)(A) Equation of circle (x+5)2+(y−12)2=81 Distance between the origin and the centre of the circle =√52+122=13 Minimum value of √x2+y2  =13−r=13−9=4 (A)→(R) (B) 3x+6y=k ⇒3x+6yk=1     ⋯(1) Now, 2x2+2xy+3y2−1=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. 2−9k2+3−36k2=0⇒5−45k2=0 ∴k2=9 (B)→(U)  Suggest corrections   