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Question

Find the equation of chord of contact of circle x2+y23x+4y+5=0 w.r.t point (1,2)

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Solution

Equation of the circle is given as,
x2+y23x+4y+5=0 ………….(1)
x2+y2+2(3/2)x+2.(2)y+5=0
Where g=3/2,f=2,c=5
Equation of the chord of contact is given by
x1x+y1y+g(x1+x)+f(y1+y)+c=0
x1x+y1y+(32)(x1+x2)+2(y1+y)+5=0
Hence equation of the chord of contact of circle (1) w.r.t (1,2)
1x+2y+(32)(1+x)+2(2+y)+5=0
x+2y323x2+4+4y+5=0
2x+4y33x+8+8y+102=0
2x+4y33x+8+8y+10=0
x+12y+15=0
x12y15=0
x12y=15.

1215658_1300951_ans_fe1f22beee294c2887af7846d4b8ed3a.JPG

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