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Question

The equation of the circle described on the chord 3x+y+5=0 of the circle x2+y2=16 as diameter is :

A
x2+y2+3x+y11=0
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B
x2+y2+3x+y+1=0
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C
x2+y2+3x+y2=0
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D
x2+y2+3x+y22=0
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Solution

The correct option is A x2+y2+3x+y11=0
3x+y+5=0 is a chord of circle x2+y2=16
y=53x ....... (i)
Put the value of y in equation of circle, we get
x2+(53x)2=16
x2+9x2+25+30x=16
10x2+30x+9=0
x=30±9004(10)(9)20
x=30±90036020
x=30±54020
x=30±61520
x=15±31510
y=53[15+31510],53[1531510]
y=591510,5+91510
End points of chord are [15+31510,591510],[1531510,5+91510]
This chord acts as a diameter for another circle.
Center = mid point of above points
=[302×10,102×10]=[32,12]
Equation of circle is :(x+32)2+(y+12)2=r2 ....... (ii)
Radius = distance between centre & [15+31510,591510]
r2=[15+31510+32]2+[591510+12]2
r2=[15+315+1510]2+[5915+510]2
r2=9×15100+81×15100
r2=15100[90]=272
Put value of r2 in (ii), we get
(x+32)2+(y+12)2=272
4x2+9+12x+4y2+1+4y=54
4x2+4y2+12x+4y=44
x2+y2+3x+y11=0

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