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1

### JEE Advanced 2016 Paper 1 Offline

MCQ (More than One Correct Answer)
Let RS be the diameter of the circle $${x^2}\, + \,{y^2} = 1$$, where S is the point (1, 0). Let P be a variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. Then the locus of E passes through the point (s)
A
$$\left( {{1 \over 3}\,,{1 \over {\sqrt 3 }}} \right)$$
B
$$\left( {{1 \over 4}\,,{1 \over 2}} \right)$$
C
$$\left( {{1 \over 3}\,, - {1 \over {\sqrt 3 }}} \right)$$
D
$$\left( {{1 \over 4}\,,-{1 \over 2}} \right)$$

## Explanation

Let, P $$\equiv$$ (cos$$\theta$$, sin$$\theta$$)

$$\therefore$$ equation of tangent and normal at P

x cos$$\theta$$ + y sin$$\theta$$ = 1 ..... (1)

and y = x tan$$\theta$$ ...... (2)

Now, equation of tangent at S : x = 1 ...... (3)

Solving (1) and (3), Q $$\equiv$$ (1, cosec$$\theta$$ $$-$$ cot$$\theta$$)

$$\therefore$$ equation of straight line parallel to RS drawn from Q

y = cosec$$\theta$$ $$-$$ cot$$\theta$$ ..... (4)

Let, E $$\equiv$$ (h, k)

$$\therefore$$ k = h tan$$\theta$$ [from (2)]

or, $$\tan \theta = {k \over h}$$

Again, k = cosec$$\theta$$ $$-$$ cot$$\theta$$ [from (4)] or, $$k = {{1 - \cos \theta } \over {\sin \theta }}$$

or, $$k = {{1 - {h \over {\sqrt {{h^2} + {k^2}} }}} \over {{k \over {\sqrt {{h^2} + {k^2}} }}}} = {{\sqrt {{h^2} + {k^2}} - h} \over k}$$

or, $${k^2} = \sqrt {{h^2} + {k^2}} - h$$

or, $$h + {k^2} = \sqrt {{h^2} + {k^2}}$$

$$\therefore$$ locus of E $$x + {y^2} = \sqrt {{x^2} + {y^2}}$$

Clearly, points $$\left( {{1 \over 3},{1 \over {\sqrt 3 }}} \right)$$ and $$\left( {{1 \over 3}, - {1 \over {\sqrt 3 }}} \right)$$ are on locus of E.

Therefore, (A) and (C) are the correct options.

2

### JEE Advanced 2014 Paper 1 Offline

MCQ (More than One Correct Answer)
A circle S passes through the point (0, 1) and is orthogonal to the circles $${(x - 1)^2}\, + \,{y^2} = 16\,\,and\,\,{x^2}\, + \,{y^2} = 1$$. Then
A
B
C
centre of S is (- 7, 1)
D
centre of S is (- 8, 1)

## Explanation

Let, the equation of the required circle is

$${x^2} + {y^2} + 2gx + 2fy + c = 0$$ ..... (1)

Circle (I) cuts the circle $${(x - 1)^2} + {y^2} = 16$$

i.e., $${x^2} + {y^2} - 2x = 15$$ orthogonally

$$\therefore$$ $$2( - g + 0) = - 15 + c$$

or, $$- 2g = - 15 + c$$

The circle (1) also cuts the circle $${x^2} + {y^2} = 1$$ orthogonally.

$$\therefore$$ 0 = $$-$$1 + c or, c = 1

$$\therefore$$ g = 7

Now, the circle (1) passes through the point (0, 1).

$$\therefore$$ $$2f + 1 + c = 0$$ or, $$2f + 1 + 1 = 0$$ or, f = $$-$$1

$$\therefore$$ the equation of the required circle is

$${x^2} + {y^2} + 14x - 2y + 1 = 0$$

whose centre is ($$-$$7, 1) and radius $$= \sqrt {49 + 1 - 1} = 7$$ units

Therefore, (B) and (C) are the correct option.

Note :

The condition of the circle $${x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$$ cuts orthogonally to the circle $${x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$$ is $$2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2}$$

3

### JEE Advanced 2013 Paper 2 Offline

MCQ (More than One Correct Answer)
Circle (s) touching x-axis at a distance 3 from the origin and having an intercept of length $$2\sqrt 7$$ on y-axis is (are)
A
$${x^2}\, + \,{y^2}\, - \,6x\,\, + 8y\, + 9 = 0$$
B
$${x^2}\, + \,{y^2}\, - \,6x\,\, + 7y\, + 9 = 0$$
C
$${x^2}\, + \,{y^2}\, - \,6x\,\, - 8y\, + 9 = 0$$
D
$${x^2}\, + \,{y^2}\, - \,6x\,\,- 7y\, + 9 = 0$$
4

### IIT-JEE 1998

MCQ (More than One Correct Answer)
If the circle $${x^2}\, + \,{y^2} = \,{a^2}$$ intersects the hyperbola $$xy = {c^2}$$ in four points $$P\,({x_1},\,{y_1}),\,Q\,\,({x_2},\,{y_2}),\,\,R\,({x_3},\,{y_3}),\,S\,({x_4},\,{y_4}),$$ then
A
$${x_1}\, + \,{x_2} + \,{x_3}\, + \,{x_4}\, = 0$$
B
$${y_1}\, + \,{y_2} + \,{y_3}\, + \,{y_4}\, = 0$$
C
$${x_1}\,{x_2}\,{x_3}\,{x_4}\, = {c^4}$$
D
$${y_1}\,{y_2}\,{y_3}\,{y_4}\, = {c^4}$$

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