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1

### AIEEE 2009

If $$P$$ and $$Q$$ are the points of intersection of the circles
$${x^2} + {y^2} + 3x + 7y + 2p - 5 = 0$$ and $${x^2} + {y^2} + 2x + 2y - {p^2} = 0$$ then there is a circle passing through $$P,Q$$ and $$(1, 1)$$ for:
A
all except one value of $$p$$
B
all except two values of $$p$$
C
exactly one value of $$p$$
D
all values of $$p$$

## Explanation

The given circles are

$${S_1} \equiv {x^2} + {y^2} + 3x + 7y + 2p - 5 = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$${S_2} \equiv {x^2} + {y^2} + 2x + 2y - {p^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

$$\therefore$$ Equation of common chord $$PQ$$ is $${S_1} - {S_2} = 0$$

$$\Rightarrow L \equiv x + 5y + {p^2} + 2p - 5 = 0$$

$$\Rightarrow$$ Equation of circle passing through $$P$$ and $$Q$$ is

$${S_1} + \lambda \,\,L = 0$$

$$\Rightarrow \left( {{x^2} + {y^2} + 3x + 7y + 2p - 5} \right) + \lambda$$

$$\left( {x + 5y + {p^2} + 2p - 5} \right) = 0$$

As it passes through $$\left( {1,1} \right),$$ therefore

$$\Rightarrow \left( {7 + 2p} \right) + \lambda \left( {2p + {p^2} + 1} \right) = 0$$

$$\Rightarrow \lambda = - {{2p + 7} \over {\left( {p + 1} \right)}},$$

which does not exist for $$p=-1$$
2

### AIEEE 2008

The differential equation of the family of circles with fixed radius $$5$$ units and centre on the line $$y = 2$$ is
A
$$\left( {x - 2} \right){y^2} = 25 - {\left( {y - 2} \right)^2}$$
B
$$\left( {y - 2} \right){y^2} = 25 - {\left( {y - 2} \right)^2}$$
C
$${\left( {y - 2} \right)^2}{y^2} = 25 - {\left( {y - 2} \right)^2}$$
D
$${\left( {x - 2} \right)^2}{y^2} = 25 - {\left( {y - 2} \right)^2}$$

## Explanation

Let the center of the circle be $$(h, 2)$$

$$\therefore$$ Equation of circle is

$${\left( {x - h} \right)^2} + \left( {y - 2} \right){}^2 = 25\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

Differentiating with respect to $$x,$$ we get

$$2\left( {x - h} \right) + 2\left( {y - 2} \right){{dy} \over {dx}} = 0$$

$$\Rightarrow x - h = - \left( {y - 2} \right){{dy} \over {dx}}$$

Substituting in equation $$(1)$$ we get

$${\left( {y - 2} \right)^2}{\left( {{{dy} \over {dx}}} \right)^2} + {\left( {y - 2} \right)^2} = 25$$

$$\Rightarrow {\left( {y - 2} \right)^2}{\left( {y'} \right)^2} = 25 - {\left( {y - 2} \right)^2}$$
3

### AIEEE 2008

The point diametrically opposite to the point $$P(1, 0)$$ on the circle $${x^2} + {y^2} + 2x + 4y - 3 = 0$$ is
A
$$(3, -4)$$
B
$$(-3, 4)$$
C
$$(-3, -4)$$
D
$$(3, 4)$$

## Explanation

The given circle is $${x^2} + {y^2} + 2x + 4y - 3 = 0$$ Center $$(-1,-2)$$

Let $$Q$$ $$\left( {\alpha ,\beta } \right)$$ be the point diametrically opposite to the point $$P(1,0),$$

then $${{1 + \alpha } \over 2} = - 1$$ and $${{0 + \beta } \over 2} = - 2$$

$$\Rightarrow \alpha = - 3,\beta = - 4,$$ So, $$Q$$ is $$\left( { - 3, - 4} \right)$$
4

### AIEEE 2007

Consider a family of circles which are passing through the point $$(-1, 1)$$ and are tangent to $$x$$-axis. If $$(h, k)$$ are the coordinate of the centre of the circles, then the set of values of $$k$$ is given by the interval
A
$$- {1 \over 2} \le k \le {1 \over 2}$$
B
$$k \le {1 \over 2}$$
C
$$0 \le k \le {1 \over 2}$$
D
$$k \ge {1 \over 2}$$

## Explanation

Equation of circle whose center is $$\left( {h,k} \right)$$

i.e $${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {k^2}$$ (radius of circle $$=k$$ because circle is tangent to $$x$$-axis)

Equation of circle passing through $$\left( { - 1, + 1} \right)$$

$$\therefore$$ $${\left( { - 1, - h} \right)^2} + {\left( {1 - k} \right)^2} = {k^2}$$

$$\Rightarrow 1 + {h^2} + 2h + 1 + {k^2} - 2k = {k^2}$$

$$\Rightarrow {h^2} + 2h - 2k + 2 = 0$$

$$D \ge 0$$

$$\therefore$$ $${\left( 2 \right)^2} - 4 \times 1.\left( { - 2k + 2} \right) \ge 0$$

$$\Rightarrow 4 - 4\left( { - 2k + 2} \right) \ge 0$$

$$\Rightarrow 1 + 2k - 2 \ge 0$$

$$\Rightarrow k \ge {1 \over 2}$$

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