1

IIT-JEE 1996

MCQ (Single Correct Answer)
The angle between a pair of tangents drawn from a point P to the circle $${x^2}\, + \,{y^2}\, + \,\,4x\, - \,6\,y\, + \,9\,{\sin ^2}\,\alpha \, + \,13\,{\cos ^2}\,\alpha \, = \,0$$ is $$2\,\alpha $$.
The equation of the locus of the point P is
A
$${x^2}\, + \,{y^2}\, + \,\,4x\, - \,6\,y\, + \,4\, = \,0$$
B
$${x^2}\, + \,{y^2}\, + \,\,4x\, - \,6\,y\,\, - \,9\,\, = \,0$$
C
$${x^2}\, + \,{y^2}\, + \,\,4x\, - \,6\,y\,\, - \,4\,\, = \,0$$
D
$${x^2}\, + \,{y^2}\, + \,\,4x\, - \,6\,y\,\, + \,9\,\, = \,0$$
2

IIT-JEE 1994

MCQ (Single Correct Answer)
The circles $${x^2} + {y^2} - 10x + 16 = 0$$ and $${x^2} + {y^2} = {r^2}$$ intersect each other in two distinct points if
A
r < 2
B
r > 8
C
2 < r < 8
D
$$2 \le r \le 8$$
3

IIT-JEE 1993

MCQ (Single Correct Answer)
The locus of the centre of a circle, which touches externally the circle $${x^2} + {y^2} - 6x - 6y + 14 = 0$$ and also touches the y-axis, is given by the equation:
A
$${x^2} - 6x - 10y + 14 = 0$$
B
$${x^2} - 10x - 6y + 14 = 0$$
C
$${y^2} - 6x - 10y + 14 = 0$$
D
$${y^2} - 10x - 6y + 14 = 0$$
4

IIT-JEE 1992

MCQ (Single Correct Answer)
The centre of a circle passing through the points (0, 0), (1, 0) and touching the circle $${x^2} + {y^2} = 9$$is
A
$$\left( {{3 \over 2},{1 \over 2}} \right)\,$$
B
$$\left( {{1 \over 2},{3 \over 2}} \right)\,$$
C
$$\left( {{1 \over 2},{1 \over 2}} \right)\,$$
D
$$\left( {{1 \over 2}, - {2^{{1 \over 2}}}} \right)\,$$

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