WB JEE 2010 | Circle Question 8 | Mathematics

WB JEE 2010

MCQ (Single Correct Answer)

-0.25

The straight line x + y $$-$$ 1 = 0 meets the circle x² + y² $$-$$ 6x $$-$$ 8y = 0 at A and B. Then the equation of the circle of which AB is a diameter is

$${x^2} + {y^2} - 2y - 6 = 0$$

$${x^2} + {y^2} + 2y - 6 = 0$$

$$2({x^2} + {y^2}) + 2y - 6 = 0$$

$$3({x^2} + {y^2}) + 2y - 6 = 0$$

WB JEE 2011

MCQ (Single Correct Answer)

-0.25

If the straight line y = mx lies outside of the circle x² + y² $$-$$ 20y + 90 = 0, then the value of m will satisfy

m < 3

| m | < 3

m > 3

| m | > 3

WB JEE 2011

MCQ (Single Correct Answer)

-0.25

The locus of the centre of a circle which passes through two variable points (a, 0), ($$-$$a, 0) is

x = 1

x + y = a

x + y = 2a

x = 0

WB JEE 2011

MCQ (Single Correct Answer)

-0.25

The intercept on the line y = x by the circle x² + y² $$-$$ 2x = 0 is AB. Equation of the circle with AB as diameter is

x² + y² = 1

x(x $$-$$ 1) + y(y $$-$$ 1) = 0

x² + y² = 2

(x $$-$$ 1) (x $$-$$ 2) + (y $$-$$ 1) (y $$-$$ 2) = 0

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