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Question

Let f(x,y)=0 be the equation of a circle such that f(0,y)=0 has equal real roots and f(x,0)=0 has two distinct real roots. Let g(x,y)=0 be the locus of points P from where tangents to circle f(x,y)=0 make angle π3 between them and g(x,y)=x2+y25x4y+c,cR. Let Q be a point from where tangents drawn to circle g(x,y)=0 are mutually perpendicular. If A,B are the points of contact of tangents drawn from Q to circle g(x,y)=0, then area of triangle QAB is

A
2512
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B
258
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C
254
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D
252
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Solution

The correct option is D 252
Given f(x,y)=0 is circle.
As f(0,y) has equal roots hence f(x,y)=0 touches the yaxis and as f(x,0)=0 has two distinct real roots hence f(x,y)=0 cuts the xaxis in two distinct points.


Given g(x,y)=x2+y25x4y+c
centre=(52,2), radius=254+4c

From figure, radius of g(x,y)= twice the radius of f(x,y)=0 and r=52
Radius of g(x,y)=5
Hence 254+4c=25c=594

Equation of g(x,y) is
x2+y25x4y594=0

Area of ΔQAB=12×5×5=252

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