1
WB JEE 2019
MCQ (Single Correct Answer)
+1
-0.25
Change Language
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
A
$$\left( {{1 \over 2},{1 \over 4}} \right)$$
B
$$\left( {{1 \over 2},{{\sqrt 3 } \over 4}} \right)$$
C
$$\left( {{1 \over 2},{1 \over {2\sqrt 3 }}} \right)$$
D
$$\left( {{1 \over 2},{{ - 1} \over {\sqrt 3 }}} \right)$$
2
WB JEE 2018
MCQ (Single Correct Answer)
+1
-0.25
Change Language
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
A
x2 + y2 + 4 = 0
B
x2 + y2 = 4
C
x2 + y2 $$-$$ 8x $$-$$ 12y + 48 = 0
D
x2 + y2 = 9
3
WB JEE 2018
MCQ (Single Correct Answer)
+1
-0.25
Change Language
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
A
x2 + y2 + 4x + 6y + 9 = 0
B
x2 + y2 $$-$$ 4x + 6y + 9 = 0
C
x2 + y2 $$-$$ 4x $$-$$ 6y + 9 = 0
D
x2 + y2 + 4x $$-$$ 6y + 9 = 0
4
WB JEE 2018
MCQ (Single Correct Answer)
+1
-0.25
Change Language
If one of the diameter of the circle, given by the equation x2 + y2 + 4x + 6y $$-$$ 12 = 0, is a chord of a circle S, whose centre is (2, $$-$$ 3), the radius of S is
A
$$\sqrt {41} $$ unit
B
3$$\sqrt {5} $$ unit
C
5$$\sqrt {2} $$ unit
D
2$$\sqrt {5} $$ unit
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