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1

JEE Main 2015 (Offline)

The number of common tangents to the circles $${x^2} + {y^2} - 4x - 6x - 12 = 0$$ and $${x^2} + {y^2} + 6x + 18y + 26 = 0,$$ is :
A
$$3$$
B
$$4$$
C
$$1$$
D
$$2$$

Explanation

$${x^2} + {y^2} - 4x - 6y - 12 = 0\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

Center, $${c_1} = \left( {2,\,3} \right)$$ and Radius, $${r_1} = 5$$ units

$${x^2} + {y^2} + 6x + 18y + 26 = 0\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Center, $${c_2} = \left( { - 3, - 9} \right)$$ and Radius, $${r_2} = 8$$ units

$${C_1}{C_2} = \sqrt {{{\left( {2 + 3} \right)}^2} + {{\left( {3 + 9} \right)}^2}} = 13\,\,$$ units

$${r_1} + {r_2} = 5 + 8 = 13$$

$$\therefore$$ $${C_1}{C_2} = {r_1} + {r_2}$$

Therefore there are three common tangents.
2

JEE Main 2015 (Offline)

Locus of the image of the point $$(2, 3)$$ in the line $$\left( {2x - 3y + 4} \right) + k\left( {x - 2y + 3} \right) = 0,\,k \in R,$$ is a:
A
circle of radius $$\sqrt 2$$.
B
circle of radius $$\sqrt 3$$.
C
straight line parallel to $$x$$-axis
D
straight line parallel to $$y$$-axis

Explanation

Intersection point of $$2x - 3y + 4 = 0$$

and $$x-2y+3=0$$ is $$(1, 2)$$

Since, $$P$$ is the fixed point for given family of lines

So, $$PB=PA$$

$${\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = {\left( {2 - 1} \right)^2} + {\left( {3 - 2} \right)^2}$$

$${\left( {\alpha - 1} \right)^2} + {\left( {\beta - 2} \right)^2} = 1 + 1 = 2$$

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

$${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {r^2}$$

Therefore, given locus is a circle with center $$(1, 2)$$ and radius $$\sqrt 2 .$$
3

JEE Main 2014 (Offline)

Let $$C$$ be the circle with centre at $$(1, 1)$$ and radius $$=$$ $$1$$. If $$T$$ is the circle centred at $$(0, y)$$, passing through origin and touching the circle $$C$$ externally, then the radius of $$T$$ is equal to
A
$${1 \over 2}$$
B
$${1 \over 4}$$
C
$${{\sqrt 3 } \over {\sqrt 2 }}$$
D
$${{\sqrt 3 } \over 2}$$

Explanation

Equation of circle $$C \equiv {\left( {x - 1} \right)^2} + {\left( {y - 1} \right)^2} = 1$$

Radius of $$T = \left| y \right|$$

$$T$$ touches $$C$$ externally

therefore,

Distance between the centers $$=$$ sum of their radii

$$\Rightarrow \sqrt {{{\left( {0 - 1} \right)}^2} + {{\left( {y - 1} \right)}^2}} = 1 + \left| y \right|$$

$$\Rightarrow {\left( {0 - 1} \right)^2} + {\left( {y - 1} \right)^2} = {\left( {1 + \left| y \right|} \right)^2}$$

$$\Rightarrow 1 + {y^2} + 1 - 2y = 1 + {y^2} + 2\left| y \right|$$

$$2\left| y \right| = 1 - 2y$$

If $$y>0$$ then $$2y=1-2y$$ $$\Rightarrow y = {1 \over 4}$$

$$y<0$$ then $$-2y=1-2y$$ $$\Rightarrow 0 = 1$$ (not possible)

$$\therefore$$ $$y = {1 \over 4}$$
4

JEE Main 2013 (Offline)

The circle passing through $$(1, -2)$$ and touching the axis of $$x$$ at $$(3, 0)$$ also passes through the point
A
$$\left( { - 5,\,2} \right)$$
B
$$\left( { 2,\,-5} \right)$$
C
$$\left( { 5,\,-2} \right)$$
D
$$\left( { - 2,\,5} \right)$$

Explanation

Since circle touches $$x$$-axis at $$(3,0)$$

$$\therefore$$ The equation of circle be

$${\left( {x - 3} \right)^2} + {\left( {y - 0} \right)^2} + \lambda y = 0$$

As it passes through $$(1, -2)$$

$$\therefore$$ Put $$x=1,$$ $$y=-2$$

$$\Rightarrow {\left( {1 - 3} \right)^2} + {\left( { - 2} \right)^2} + \lambda \left( { - 2} \right) = 0$$

$$\Rightarrow \lambda = 4$$

$$\therefore$$ equation of circle is $${\left( {x - 3} \right)^2} + {y^2} - 8 = 0$$

Now, from the options $$\left( {5, - 2} \right)$$ satisfies equation of circle.

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