IIT-JEE 1993 | Circle Question 59 | Mathematics

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:

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

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

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

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

IIT-JEE 1992

MCQ (Single Correct Answer)

-0.5

The centre of a circle passing through the points (0, 0), (1, 0) and touching the circle $${x^2} + {y^2} = 9$$is

$$\left( {{3 \over 2},{1 \over 2}} \right)\,$$

$$\left( {{1 \over 2},{3 \over 2}} \right)\,$$

$$\left( {{1 \over 2},{1 \over 2}} \right)\,$$

$$\left( {{1 \over 2}, - {2^{{1 \over 2}}}} \right)\,$$

IIT-JEE 1989

MCQ (Single Correct Answer)

-0.5

The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle of area 154 sq. units. Then the equation of this circle is

$${x^2} + {y^2} + 2x - 2y = 62$$

$${x^2} + {y^2} + 2x - 2y = 47$$

$${x^2} + {y^2} - 2x + 2y = 47$$

$${x^2} + {y^2} - 2x + 2y = 62$$c

IIT-JEE 1989

MCQ (Single Correct Answer)

-0.5

If the two circles $${(x - 1)^2} + {(y - 3)^2} = {r^2}$$ and $${x^2} + {y^2} - 8x + 2y + 8 = 0$$ intersect in two distinct points, then

2 < r < 8

r < 2

r = 2

r > 2

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