WB JEE 2019 | Circle Question 25 | Mathematics

WB JEE 2019

MCQ (Single Correct Answer)

-0.25

A variable circle passes through the fixed point A(p, q) and touches X-axis. The locus of the other end of the diameter through A is

$${(x - p)^2} = 4qy$$

$${(x - q)^2} = 4py$$

$${(y - p)^2} = 4qx$$

$${(y - q)^2} = 4px$$

WB JEE 2019

MCQ (Single Correct Answer)

-0.25

If P(0, 0), Q(1, 0) and R$$\left( {{1 \over 2},{{\sqrt 3 } \over 2}} \right)$$ are three given points, then the centre of the circle for which the lines PQ, QR and RP are the tangents is

$$\left( {{1 \over 2},{1 \over 4}} \right)$$

$$\left( {{1 \over 2},{{\sqrt 3 } \over 4}} \right)$$

$$\left( {{1 \over 2},{1 \over {2\sqrt 3 }}} \right)$$

$$\left( {{1 \over 2},{{ - 1} \over {\sqrt 3 }}} \right)$$

WB JEE 2018

MCQ (Single Correct Answer)

-0.25

Without changing the direction of the axes, the origin is transferred to the point (2, 3). Then the equation x² + y² $$-$$ 4x $$-$$ 6y + 9 = 0 changes to

x² + y² + 4 = 0

x² + y² = 4

x² + y² $$-$$ 8x $$-$$ 12y + 48 = 0

x² + y² = 9

WB JEE 2018

MCQ (Single Correct Answer)

-0.25

The angle between a pair of tangents drawn from a point P to the circle x² + y² + 4x $$-$$ 6y + 9sin²$$\alpha$$ + 13cos²$$\alpha$$ = 0 is 2$$\alpha$$. The equation of the locus of the point P is

x² + y² + 4x + 6y + 9 = 0

x² + y² $$-$$ 4x + 6y + 9 = 0

x² + y² $$-$$ 4x $$-$$ 6y + 9 = 0

x² + y² + 4x $$-$$ 6y + 9 = 0

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