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JEE Mains Previous Years Questions with Solutions

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1

AIEEE 2004

MCQ (Single Correct Answer)
If a circle passes through the point (a, b) and cuts the circle $${x^2}\, + \,{y^2} = 4$$ orthogonally, then the locus of its centre is
A
$$2ax\, - 2by\, - ({a^2}\, + \,{b^2} + 4) = 0$$
B
$$2ax\, + 2by\, - ({a^2}\, + \,{b^2} + 4) = 0$$
C
$$2ax\, - 2by\, + ({a^2}\, + \,{b^2} + 4) = 0$$
D
$$2ax\, + 2by\, + ({a^2}\, + \,{b^2} + 4) = 0$$

Explanation

Let the variable circle is

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

It passes through $$(a,b)$$

$$\therefore$$ $${a^2} + {b^2} + 2ga + 2fb + c = 0\,\,\,\,\,\,\,...\left( 2 \right)$$

$$(1)$$ cuts $${x^2} + {y^2} = 4$$ orthogonally

$$\therefore$$ $$2\left( {g \times 0 + f \times 0} \right) = c - 4 \Rightarrow c = 4$$

$$\therefore$$ from $$(2)$$ $$\,\,\,{a^2} + {b^2} + 2ga + 2fb + 4 = 0$$

$$\therefore$$ Locus of center $$\left( { - g, - f} \right)$$ is

$${a^2} + {b^2} - 2ax - 2by + 4 = 0$$

or $$2ax + 2by = {a^2} + {b^2} + 4$$
2

AIEEE 2004

MCQ (Single Correct Answer)
If the lines 2x + 3y + 1 + 0 and 3x - y - 4 = 0 lie along diameter of a circle of circumference $$10\,\pi $$, then the equation of the circle is
A
$${x^2}\, + \,{y^2} + \,2x\, - \,2y - \,23\,\, = 0$$
B
$${x^2}\, + \,{y^2} - \,2x\, - \,2y - \,23\,\, = 0$$
C
$${x^2}\, + \,{y^2} + \,2x\, + \,2y - \,23\,\, = 0$$
D
$${x^2}\, + \,{y^2} - \,2x\, + \,2y - \,23\,\, = 0$$

Explanation

Two diameters are along

$$2x+3y+1=0$$ and $$3x-y-4=0$$

solving we get center $$(1,-1)$$

circumference $$ = 2\pi r = 10\pi $$

$$\therefore$$ $$r=5$$.

Required circle is, $${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {5^2}$$

$$ \Rightarrow {x^2} + {y^2} - 2x + 2y - 23 = 0$$
3

AIEEE 2003

MCQ (Single Correct Answer)
The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle having area as 154 sq. units. Then the equation of the circle is
A
$${x^2}\, + \,{y^2} - \,2x\, + \,2y\,\, = \,62$$
B
$${x^2}\, + \,{y^2} + \,2x\, - \,2y\,\, = \,62$$
C
$${x^2}\, + \,{y^2} + \,2x\, - \,2y\,\, = \,47$$
D
$${x^2}\, + \,{y^2} - \,2x\, + \,2y\,\, = \,47$$

Explanation

$$\pi {r^2} = 154 \Rightarrow r = 7$$

For center on solving equation

$$2x - 3y = 5\& 3x - 4y = 7$$

we get $$x = 1,\,y = - 1$$

$$\therefore$$ center $$=(1,-1)$$

Equation of circle,

$${\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {7^2}$$

$${x^2} + {y^2} - 2x + 2y = 47$$
4

AIEEE 2003

MCQ (Single Correct Answer)
If the two circles $${(x - 1)^2}\, + \,{(y - 3)^2} = \,{r^2}$$ and $$\,{x^2}\, + \,{y^2} - \,8x\, + \,2y\, + \,\,8\,\, = 0$$ intersect in two distinct point, then
A
$$r > 2$$
B
$$2 < r < 8$$
C
$$r < 2$$
D
$$r = 2.$$

Explanation

$$\left| {{r_1} - {r_2}} \right| < {C_1}{C_2}$$ for intersection

$$ \Rightarrow r - 3 < 5 \Rightarrow r < 8\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

and $${r_1} + {r_2} > {C_1}{C_2},\,$$

$$r + 3 > 5 \Rightarrow r > 2\,\,\,...\left( 2 \right)$$

From $$\left( 1 \right)$$ and $$\left( 2 \right),$$ $$2 < r < 8.$$

Questions Asked from Circle

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