IIT-JEE 1994 | Circle Question 58 | Mathematics

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:

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

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

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

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

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

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