IIT-JEE 1992 | Circle Question 64 | Mathematics

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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

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

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

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

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

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

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

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