IIT-JEE 1984 | Circle Question 85 | Mathematics

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IIT-JEE 1984

MCQ (Single Correct Answer)

-0.5

The locus of the mid-point of a chord of the circle $${x^2} + {y^2} = 4$$ which subtends a right angle at the origin is

x + y = 2

$${x^2} + {y^2} = 1$$

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

$$x + y $$=1

IIT-JEE 1983

MCQ (Single Correct Answer)

-0.25

The centre of the circle passing through the point (0, 1) and touching the curve $$\,y = {x^2}$$ at (2, 4) is

$$\left( {{{ - 16} \over 5},{{ - 27} \over {10}}} \right)$$

$$\left( {{{ - 16} \over 7},{{53} \over {10}}} \right)$$

$$\left( {{{ - 16} \over 5},{{53} \over {10}}} \right)$$

none of these

IIT-JEE 1983

MCQ (Single Correct Answer)

-0.25

The equation of the circle passing through (1, 1) and the points of intersection of $${x^2} + {y^2} + 13x - 3y = 0$$ and $$2{x^2} + 2{y^2} + 4x - 7y - 25 = 0$$ is

$$4{x^2} + 4{y^2} - 30x - 10y - 25 = 0$$

$$4{x^2} + 4{y^2} + 30x - 13y - 25 = 0$$

$$4{x^2} + 4{y^2} - 17x - 10y + 25 = 0$$

none of these

IIT-JEE 1980

MCQ (Single Correct Answer)

-0.25

Two circles $${x^2} + {y^2} = 6$$ and $${x^2} + {y^2} - 6x + 8 = 0$$ are given. Then the equation of the circle through their points of intersection and the point (1, 1) is

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

$${x^2} + {y^2} - 3x + 1 = 0$$

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

none of these

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