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Question

Match the following by approximately matching the lists based on the information given in Column I and Column II

Column 1Column 2a. The length of the common chord of two circles of radii 3 and p. 14 units which intersect orthogonally is k5,then k is equal to b. The circumference of the circle x2+y2+4x+12y+p=0 q. 24 is bisected by the circle x2+y22x+8yq=0, then p+q is equal to c. Number of distinct chords of the circle 2x(x2)+y(2y1)r. 32=0 chords are passing through the point (2,12) and are bisected on x-axis is d. One of the diameters of the circle circumscribing the rectangle s. 36ABCD is 4y=x+7. If A and B are the points (3,4) and (5,4) respectively, then the area of rectangle is

A
aq, bs, cp, dr
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B
ap, bs, cq, dr
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C
aq, br, cp, ds
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D
ar, bs, cp, dq
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Solution

The correct option is A aq, bs, cp, dr
(a)
Let the length of the common chord be 2a. Then, 9a2+16a2=5
or 16a2=59a2
or 16a2=25+9a2109a2
or 109a2=18
or 100(9a2)=324
or 100a2=576
or a=576100=2410
2a=245=k5k=24

(b)
The equation of common chord is 6x+4y+p+q=0
The common chord passes through the center (2,6) of the circle x2+y2+4x+12y+p=0. Therefore, p+q=36

(c)
The equation of the circle is 2x2+2y222xy=0
let (α,0) be the midpoint of a chord. Then, the equation of the chord is 2αx2(x+α)12(y+0)=2α222α
Since it passes through the point (2,12), we have 22α2(2+α)14=2α222α
8α2122α+9=0
(22α3)2=0
α=322,322
Therefore, the number of chords is 1

(d)
Midpoint of AB(1,4)
The equation perpenduicular bisector of AB is x=1
A dimeter of the circle is 4y=x+7
Therefore, the center of the circle is (1,2)
Hence, the sides of the rectangle are 8 and 4. Therefore,
Area =32

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