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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

If the equation of one tangent to the circle with centre at (2, $$-$$1) from the origin is 3x + y = 0, then the equation of the other tangent through the origin is

A
$$3x - y = 0$$
B
$$x + 3y = 0$$
C
$$x - 3y = 0$$
D
$$x + 2y = 0$$

একটি বৃত্তের কেন্দ্র (2, $$-$$1) দেওয়া আছে। ঐ বৃত্তের মূলবিন্দু থেকে অঙ্কিত একটি স্পর্শকের সমীকরণ হল $$3x + y = 0$$ । সেক্ষেত্রে মূলবিন্দু থেকে অঙ্কিত অপর স্পর্শকের সমীকরণ হবে

A
$$3x - y = 0$$
B
$$x + 3y = 0$$
C
$$x - 3y = 0$$
D
$$x + 2y = 0$$
2

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

A curve passes through the point (3, 2) for which the segment of the tangent line contained between the co-ordinate axes is bisected at the point of contact. The equation of the curve is

A
$$y = {x^2} - 7$$
B
$$x = {{{y^2}} \over 2} + 2$$
C
$$xy = 6$$
D
$${x^2} + {y^2} - 5x + 7y + 11 = 0$$

Explanation

According to the question, p(x, y) is the midpoint of line AB.

$$\therefore$$ $${{\alpha + 0} \over 2} = x \Rightarrow \alpha = 2x$$

$${{\beta + 0} \over 2} = y \Rightarrow \beta = 2y$$

$$\therefore$$ Point A = (2x, 0)

and Point B = (0, 2y)

Slope of the tangent,

$${{dy} \over {dx}} = {{2y - 0} \over {0 - 2x}}$$

$$ \Rightarrow {{dy} \over {dx}} = - {y \over x}$$

$$ \Rightarrow x\,dy + y\,dx = 0$$

$$ \Rightarrow d(xy) = 0$$

$$ \Rightarrow xy = c$$

This curve goes through point (3, 2). So this point satisfy the equation.

$$\therefore$$ 3 . 2 = c

$$\Rightarrow$$ c = 6

$$\therefore$$ Equation of the curve,

xy = 6

একটি বক্ররেখা (3, 2) বিন্দুগামী, বক্ররেখাটির একটি বিন্দুতে অঙ্কিত স্পর্শকের অক্ষদ্বয়ের মধ্যেকার ছেদিতাংশ ঐ স্পর্শবিন্দুতে সমদ্বিখন্ডিত হয়। বক্ররেখাটির সমীকরণ হবে

A
$$y = {x^2} - 7$$
B
$$x = {{{y^2}} \over 2} + 2$$
C
$$xy = 6$$
D
$${x^2} + {y^2} - 5x + 7y + 11 = 0$$
3

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
Two tangents to the circle x2 + y2 = 4 at the points A and B meet at M($$-$$4, 0). The area of the quadrilateral MAOB, where O is the origin is
A
$$4\sqrt 3 $$ sq. units
B
$$2\sqrt 3 $$ sq. units
C
$$\sqrt 3 $$ sq. units
D
$$3\sqrt 3 $$ sq. units

Explanation


Given, equation of circle x2 + y2 = 4

Centre = (0, 0) and radius = 2

$$\therefore$$ $$MA = \sqrt {{{(4)}^2} - {{(2)}^2}} = \sqrt {12} = 2\sqrt 3 $$

$$\therefore$$ Area of quadrilateral MAOB

= 2 $$\times$$ area of $$\Delta$$MAO

$$ = 2 \times {1 \over 2}$$ $$\times$$ MA $$\times$$ OA

$$ = 2\sqrt 3 $$ $$\times$$ 2

$$ = 4\sqrt 3 $$ sq. units.
x2 + y2 = 4 বৃত্তের উপরিস্থ A ও B বিন্দুতে অঙ্কিত স্পর্শকদ্বয় M($$-$$4, 0) বিন্দুতে মিলিত হয়। চতুর্ভূজ MAOB (O মূলবিন্দু) -এর ক্ষেত্রফল হবে
A
$$4\sqrt 3 $$ বর্গ একক
B
$$2\sqrt 3 $$ বর্গ একক
C
$$\sqrt 3 $$ বর্গ একক
D
$$3\sqrt 3 $$ বর্গ একক

Explanation


দেওয়া হয়েছে, বৃত্তের সমীকরণ x2 + y2 = 4

কেন্দ্র = (0, 0) ও ব্যাসার্ধ = 2

$$\therefore$$ $$MA = \sqrt {{{(4)}^2} - {{(2)}^2}} = \sqrt {12} = 2\sqrt 3 $$

$$\therefore$$ চতুর্ভুজ MAOB এর ক্ষেত্রফল

= 2 $$\times$$ $$\Delta$$MAO এর ক্ষেত্রফল

$$ = 2 \times {1 \over 2}$$ $$\times$$ MA $$\times$$ OA

$$ = 2\sqrt 3 $$ $$\times$$ 2

$$ = 4\sqrt 3 $$ বর্গ একক
4

WB JEE 2020

MCQ (Single Correct Answer)
English
Bengali
The equation of circle of radius $$\sqrt {17} $$ unit, with centre on the positive side of X-axis and through the point (0, 1) is
A
$${x^2} + {y^2} - 8x - 1 = 0$$
B
$${x^2} + {y^2} + 8x - 1 = 0$$
C
$${x^2} + {y^2} - 9y + 1 = 0$$
D
$$2{x^2} + 2{y^2} - 3x + 2y = 4$$

Explanation

Given, the radius of circle = $$\sqrt {17} $$

With centre on positive sides of X-axis

$$ \therefore $$ Equation of circle is $${(x - a)^2} + {y^2} = 17$$

Since, it passes through (0, 1)

$$ \therefore $$ a2 + 1 = 17

$$ \Rightarrow $$ a = 4

$$ \therefore $$ Equation of circle is

$${x^2} + {y^2} - 8x - 1 = 0$$

যে বৃত্ত (0, 1) বিন্দুগামী, যার ব্যাসার্ধ $$\sqrt{17}$$ একক এবং যার কেন্দ্র x-অক্ষের ধনাত্মক দিকের উপরিস্থ, তার সমীকরণ হল -

A
$${x^2} + {y^2} - 8x - 1 = 0$$
B
$${x^2} + {y^2} + 8x - 1 = 0$$
C
$${x^2} + {y^2} - 9y + 1 = 0$$
D
$$2{x^2} + 2{y^2} - 3x - 2y = 4$$

Explanation

বৃত্তের সমীকরণ $${(x - a)^2} + {(y - b)^2} = 17$$ যেখানে কেন্দ্র (a, b)

ইহা (0, 1) বিন্দুগামী

$$\therefore$$ $${a^2} + {(1 - b)^2} = 17 = {4^2} + {1^2}$$

$$\therefore$$ $$a = \pm \,4,\,1 - b = \pm \,1$$

বা, $$a = \pm \,4,\,b = 1 \mp \,1 = 0,2$$

$$\therefore$$ বৃত্তের সমীকরণ $${(x - 4)^2} + {(y - 0)^2} = 17$$

বা, $${x^2} + {y^2} - 8x - 1 = 0$$

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