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Column 1Column 2a. If ax+by5=0 is the equation of the chord of the circle    p. 6(x3)2+(y4)2=4,which passes through (2,3) and at   the greatest distance from the centre, then |a+b| is equal tob. Let O be the origin and P be a variable point on the circle     q. 3x2+y2+2x+2y=0. If the locus of midpoint of OP is    x2+y2+2gx+2fy+c=0,then(g+f)is equal to            c. The x–coordinate of the centre of the smallest circle which    r. 2   cuts the circles x2+y22x4y4=0 and x2+y210x+12y+52=0 orthogonally is                                                 d. If θ be the angle between two tangents which are drawn        s. 1   to the circlesx2+y263x6y+27=0 from the origin, then 23tanθ equals to                                                           

Which of the following is correct?
  1. as, br, cp, dq
  2. ar, bs, cp, dq
  3. as, br, cq, dp
  4. ar, bs, cq, dp


Solution

The correct option is D ar, bs, cq, dp
a.
We know that (2,3) lies inside the circle,
The chord which is at the greatest distance from the centre is bisected at (2,3), 

Slope of line joining point (2,3) and centre is 
m=4332=1
The chord will be perpendicular to this, so th slope of the chord
=1
Equation of chord is
y3=(x2)x+y5=0 
Comparing with ax+by5=0, we get
a1=b1=55a=b=1|a+b|=2

b.
Let P be the point (α,β)
Assuming the midpoint be M(h,k), so
(h,k)=(α2,β2)α=2h,β=2k 
Then P=(2h,2k)
So putting the point P in the equation of circle, we get
4h2+4k2+4h+4k=0h2+k2+h+k=0
Therefore, the locus of the midpoint is 
x2+y2+x+y=0
Comparing with x2+y2+2gx+2fy+c=0, we get
f=g=12,c=0f+g=1

c.
Given circles are 
x2+y22x4y4=0
C1=(1,2),  r1=3
x2+y210x+12y+52=0
C2=(5,6),  r2=3

The equation of line joinig C1C2 is
y2=6251(x1)y2=2x+22x+y4=0(1)
The equation of the radical axis is
8x16y56=0x2y7=0(2)
Finding the point of intersection of (1) and (2), we get
5x15=0x=3
Hence, the required x coordinate is 3.

d.
Given circle is
x2+y263x6y+27=0
C=(33,3),  r=3
Distance between centre and origin 
d=27+9=6
Now, we know that
sinθ2=rdsinθ2=36=12θ2=π6θ=π3tanθ=323tanθ=6

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