IIT-JEE 1999 | Circle Question 47 | Mathematics

IIT-JEE 1999

MCQ (Single Correct Answer)

-0.5

If two distinct chords, drawn from the point (p, q) on the circle $${x^2}\, + \,{y^2} = \,px\, + \,qy\,\,(\,where\,pq\, \ne \,0)$$ are bisected by the x - axis, then

$${p^2}\, = \,\,{q^2}$$

$$\,{p^2}\, = \,\,8\,{q^2}$$

$${p^2}\, < \,\,8\,{q^2}$$

$${p^2}\, > \,\,8\,{q^2}$$.

IIT-JEE 1996

MCQ (Single Correct Answer)

-0.25

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

$${x^2}\, + \,{y^2}\, + \,\,4x\, - \,6\,y\, + \,4\, = \,0$$

$${x^2}\, + \,{y^2}\, + \,\,4x\, - \,6\,y\,\, - \,9\,\, = \,0$$

$${x^2}\, + \,{y^2}\, + \,\,4x\, - \,6\,y\,\, - \,4\,\, = \,0$$

$${x^2}\, + \,{y^2}\, + \,\,4x\, - \,6\,y\,\, + \,9\,\, = \,0$$

IIT-JEE 1994

MCQ (Single Correct Answer)

-0.25

The circles $${x^2} + {y^2} - 10x + 16 = 0$$ and $${x^2} + {y^2} = {r^2}$$ intersect each other in two distinct points if

r < 2

r > 8

2 < r < 8

$$2 \le r \le 8$$

IIT-JEE 1993

MCQ (Single Correct Answer)

-0.25

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:

$${x^2} - 6x - 10y + 14 = 0$$

$${x^2} - 10x - 6y + 14 = 0$$

$${y^2} - 6x - 10y + 14 = 0$$

$${y^2} - 10x - 6y + 14 = 0$$

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