IIT-JEE 1989 | Circle Question 68 | Mathematics

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 1988

MCQ (Single Correct Answer)

-0.25

If a circle passes through the point (a, b) and cuts the circle $${x^2}\, + \,{y^2}\, = \,{k^2}$$ orthogonally, then the equation of the locus of its centre is

$$2\,ax\, + \,2\,by\, - \,({a^2}\, + \,{b^2}\, + \,\,{k^2})\, = \,0$$

$$2\,ax\, + \,2\,by\, - \,({a^2}\, - \,\,{b^2}\, + \,\,{k^2})\, = \,0$$

$${x^2}\, + \,{y^2}\, - \,3\,\,ax\, + \,4\,by\, + \,\,({a^2}\, + \,\,{b^2}\, - \,\,{k^2})\, = \,0$$

$${x^2}\, + \,{y^2}\, - \,2\,\,ax\, - \,4\,by\, + \,\,({a^2}\, - \,\,{b^2}\, - \,\,{k^2})\, = \,0$$.

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

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