WB JEE 2008 | Circle Question 1 | Mathematics

WB JEE 2008

MCQ (Single Correct Answer)

-0.25

The equation of the circle which passes through the points of intersection of the circles $${x^2} + {y^2} - 6x = 0$$ and $${x^2} + {y^2} - 6y = 0$$, and has its centre at $$\left( {{3 \over 2},{3 \over 2}} \right)$$ is

$${x^2} + {y^2} + 3x + 3y + 9 = 0$$

$${x^2} + {y^2} + 3x + 3y = 0$$

$${x^2} + {y^2} - 3x - 3y = 0$$

$${x^2} + {y^2} - 3x - 3y + 9 = 0$$

WB JEE 2008

MCQ (Single Correct Answer)

-0.25

The equation $$(x - {x_1})(x - {x_2}) + (y - {y_1})(y - {y_2}) = 0$$ represents a circle whose centre is

$$\left( {{{{x_1} - {x_2}} \over 2},{{{y_1} - {y_2}} \over 2}} \right)$$

$$\left( {{{{x_1} + {x_2}} \over 2},{{{y_1} + {y_2}} \over 2}} \right)$$

$$({x_1},{y_1})$$

$$({x_2},{y_2})$$

WB JEE 2008

MCQ (Single Correct Answer)

-0.25

The circles $${x^2} + {y^2} + 6x + 6y = 0$$ and $${x^2} + {y^2} - 12x - 12y = 0$$

cut orthogonally

touch each other internally

intersect in two points

touch each other externally

WB JEE 2008

MCQ (Single Correct Answer)

-0.25

The locus of the centres of the circles which touch both the axes is given by

$${x^2} - {y^2} = 0$$

$${x^2} + {y^2} = 0$$

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

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

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