IIT-JEE 1983 | Circle Question 86 | Mathematics

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

A square is inscribed in the circle $${x^2} + {y^2} - 2x + 4y + 3 = 0$$. Its sides are parallel to the coordinate axes. The one vertex of the square is

$$\left( {1 + \sqrt {2,} - 2} \right)$$

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

$$\,\,\left( {1, - 2 + \sqrt 2 } \right)$$

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

JEE Advanced Subjects