 ### JEE Mains Previous Years Questions with Solutions

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1

### 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 $$\left( { - 3, - 4} \right)$$
2

### 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 $$\left( { - 1, + 1} \right)$$

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

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

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

$$D \ge 0$$

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

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

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

$$\Rightarrow k \ge {1 \over 2}$$
3

### AIEEE 2006

Let $$C$$ be the circle with centre $$(0, 0)$$ and radius $$3$$ units. The equation of the locus of the mid points of the chords of the circle $$C$$ that subtend an angle of $${{2\pi } \over 3}$$ at its center is
A
$${x^2} + {y^2} = {3 \over 2}$$
B
$${x^2} + {y^2} = 1$$
C
$${x^2} + {y^2} = {{27} \over 4}$$
D
$${x^2} + {y^2} = {{9} \over 4}$$

## Explanation

Let $$M\left( {h,k} \right)$$ be the mid point of chord $$AB$$ where

$$\angle AOB = {{2\pi } \over 3}$$ $$\therefore$$ $$\angle AOM = {\pi \over 3}.$$ Also $$OM=$$ $$3\cos {\pi \over 3} = {3 \over 2}$$

$$\Rightarrow \sqrt {{h^2} + k{}^2} = {3 \over 2}$$

$$\Rightarrow {h^2} + {k^2} = {9 \over 4}$$

$$\therefore$$ Locus of $$\left( {h,k} \right)$$ is

$${x^2} + {y^2} = {9 \over 4}$$
4

### AIEEE 2006

If the lines $$3x - 4y - 7 = 0$$ and $$2x - 3y - 5 = 0$$ are two diameters of a circle of area $$49\pi$$ square units, the equation of the circle is
A
$$\,{x^2} + {y^2} + 2x\, - 2y - 47 = 0\,$$
B
$$\,{x^2} + {y^2} + 2x\, - 2y - 62 = 0\,$$
C
$${x^2} + {y^2} - 2x\, + 2y - 62 = 0$$
D
$${x^2} + {y^2} - 2x\, + 2y - 47 = 0$$

## Explanation

Point of intersection of $$3x - 4y - 7 = 0$$ and

$$2x - 3y - 5 = 0$$ is $$\left( {1, - 1} \right)$$ which is the center of the

circle and radius $$=7$$

$$\therefore$$ Equation is $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 49$$

$$\Rightarrow {x^2} + {y^2} - 2x + 2y - 47 = 0$$

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