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1

### 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 $$2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2}$$

2

### 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
$${x^2}\, + \,{y^2}\, - \,6x\,\,- 7y\, + 9 = 0$$
3

### IIT-JEE 1998

MCQ (More than One Correct Answer)
The number of common tangents to the circles $${x^2}\, + \,{y^2} = 4$$ and $${x^2}\, + \,{y^2}\, - 6x\, - 8y = 24$$ is
A
0
B
1
C
3
D
4
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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