IIT-JEE 2005 Screening | Circle Question 32 | Mathematics

IIT-JEE 2005 Screening

MCQ (Single Correct Answer)

-0.5

A circle is given by $${x^2}\, + \,{(y\, - \,1\,)^2}\, = \,1$$, another circle C touches it externally and also the x-axis, then thelocus of its centre is

$$\{ (x,\,y):\,\,{x^2} = \,4y\} \, \cup \,\{ (x,\,y):\,\,y \le \,0\,\} $$

$$\{ (x,\,y):\,\,{x^2} + \,{(y\, - \,1)^2}\, = \,4\} \, \cup \,\{ (x,\,\,y):\,\,y \le \,0\,\} $$

$$\{ (x,\,y):\,\,{x^2} = \,y\} \, \cup \,\{ (0,\,\,y):\,\,y \le \,0\,\} $$

$$\{ (x,\,y):\,\,{x^2} = \,4y\} \, \cup \,\{ (0,\,\,y):\,\,y \le \,0\,\} $$

IIT-JEE 2004 Screening

MCQ (Single Correct Answer)

-0.5

If one of the diameters of the circle $${x^2} + {y^2} - 2x - 6y + 6 = 0$$ is a chord to the circle with centre (2, 1), then the radius of the circle is

$${\sqrt 3 }$$

$${\sqrt 2 }$$

IIT-JEE 2003 Screening

MCQ (Single Correct Answer)

-0.5

The centre of circle inscibed in square formed by the lines $${x^2} - 8x + 12 = 0\,\,and\,{y^2} - 14y + 45 = 0$$, is

(4, 7)

(7, 4)

(9, 4)

(4, 9)

IIT-JEE 2002 Screening

MCQ (Single Correct Answer)

-0.5

If $$a > 2b > 0$$ then the positive value of $$m$$ for which $$y = mx - b\sqrt {1 + {m^2}} $$ is a common tangent to $${x^2} + {y^2} = {b^2}$$ and $${\left( {x - a} \right)^2} + {y^2} = {b^2}$$ is

$${{2b} \over {\sqrt {{a^2} - 4{b^2}} }}$$

$${{\sqrt {{a^2} - 4{b^2}} } \over {2b}}$$

$${{2b} \over {a - 2b}}$$

$${{b} \over {a - 2b}}$$

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