AIEEE 2008 | Circle Question 129 | Mathematics

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AIEEE 2008

MCQ (Single Correct Answer)

-1

The differential equation of the family of circles with fixed radius $$5$$ units and centre on the line $$y = 2$$ is

$$\left( {x - 2} \right){y^2} = 25 - {\left( {y - 2} \right)^2}$$

$$\left( {y - 2} \right){y^2} = 25 - {\left( {y - 2} \right)^2}$$

$${\left( {y - 2} \right)^2}{y^2} = 25 - {\left( {y - 2} \right)^2}$$

$${\left( {x - 2} \right)^2}{y^2} = 25 - {\left( {y - 2} \right)^2}$$

AIEEE 2007

MCQ (Single Correct Answer)

-1

Consider a family of circles which are passing through the point $$(-1, 1)$$ and are tangent to $$x$$-axis. If $$(h, k)$$ are the coordinate of the centre of the circles, then the set of values of $$k$$ is given by the interval

$$ - {1 \over 2} \le k \le {1 \over 2}$$

$$k \le {1 \over 2}$$

$$0 \le k \le {1 \over 2}$$

$$k \ge {1 \over 2}$$

AIEEE 2006

MCQ (Single Correct Answer)

-1

If the lines $$3x - 4y - 7 = 0$$ and $$2x - 3y - 5 = 0$$ are two diameters of a circle of area $$49\pi $$ square units, the equation of the circle is

$$\,{x^2} + {y^2} + 2x\, - 2y - 47 = 0\,$$

$$\,{x^2} + {y^2} + 2x\, - 2y - 62 = 0\,$$

$${x^2} + {y^2} - 2x\, + 2y - 62 = 0$$

$${x^2} + {y^2} - 2x\, + 2y - 47 = 0$$

AIEEE 2006

MCQ (Single Correct Answer)

-1

Let $$C$$ be the circle with centre $$(0, 0)$$ and radius $$3$$ units. The equation of the locus of the mid points of the chords of the circle $$C$$ that subtend an angle of $${{2\pi } \over 3}$$ at its center is

$${x^2} + {y^2} = {3 \over 2}$$

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

$${x^2} + {y^2} = {{27} \over 4}$$

$${x^2} + {y^2} = {{9} \over 4}$$

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