1
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 15th April Evening Slot

The tangent to the circle C1 : x2 + y2 $$-$$ 2x $$-$$ 1 = 0 at the point (2, 1) cuts off a chord of length 4 from a circle C2 whose center is (3, $$-$$2). The radius of C2 is :
A
2
B
$$\sqrt 2 $$
C
3
D
$$\sqrt 6 $$

Explanation

Here, equation of tangent on C1 at (2, 1) is :

2x + y $$-$$ (x + 2) $$-$$1 = 0

Or    x + y = 3

If it cuts off the chord of the circle C2 then the equation of the chord is :
x + y = 3

$$\therefore\,\,\,$$ distance of the chord from (3, $$-$$ 2) is :

d = $$\left| {{{3 - 2 - 3} \over {\sqrt 2 }}} \right|$$ = $$\sqrt 2 $$

Also, length of the chord is $$l$$ = 4

$$\therefore\,\,\,$$ radius of C2 = r = $$\sqrt {{{\left( {{l \over 2}} \right)}^2} + {d^2}} $$

= $$\sqrt {{{\left( 2 \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} = \sqrt 6 $$
2
MCQ (Single Correct Answer)

JEE Main 2018 (Online) 16th April Morning Slot

If a circle C, whose radius is 3, touches externally the circle,
$${x^2} + {y^2} + 2x - 4y - 4 = 0$$ at the point (2, 2), then the length of the intercept cut by this circle C, on the x-axis is equal to :
A
$$2\sqrt 5 $$
B
$$3\sqrt 2 $$
C
$$\sqrt 5 $$
D
$$2\sqrt 3 $$

Explanation

Given circle is :

x2 + y2 + 2x $$-$$ 4y $$-$$4 = 0

$$\therefore\,\,\,$$ its center is ($$-$$ 1, 2) and radius is 3 units.

Let A = (x, y) be the center of the circle C

$$ \therefore $$$$\,\,\,$$ $${{x - 1} \over 2}$$ = 2 $$ \Rightarrow $$ x = 5 and $${{y + 2} \over 2}$$ = 2 $$ \Rightarrow $$ y = 2

So the center of C is (5, 2) and its radius is 3

$$\therefore\,\,\,$$ Equation of center C is :

x2 + y2 $$-$$ 10x $$-$$ 4y + 20 = 0

$$\therefore\,\,\,$$ The length of the intercept it cuts on the x-axis

= 2$$\sqrt {{g^2} - c} = 2\sqrt {25 - 20} = 2\sqrt 5 $$
3
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Morning Slot

Three circles of radii a, b, c (a < b < c) touch each other externally. If they have x-axis as a common tangent, then :
A
a, b, c are in A.P.
B
$$\sqrt a ,\sqrt b ,\sqrt c $$ are in A.P
C
$${1 \over {\sqrt b }} + {1 \over {\sqrt c }}$$ = $${1 \over {\sqrt a }}$$
D
$${1 \over {\sqrt b }} = {1 \over {\sqrt a }} + {1 \over {\sqrt c }}$$

Explanation



AB = AC + CB

$$\sqrt {{{\left( {b + c} \right)}^2} - {{\left( {c - b} \right)}^2}} $$ = $$\sqrt {{{\left( {b + a} \right)}^2} - {{\left( {b - a} \right)}^2}} $$
                      + $$\sqrt {{{\left( {c + a} \right)}^2} - {{\left( {c - a} \right)}^2}} $$

$$ \Rightarrow $$  $$\sqrt {2bc} $$ = $$\sqrt {2ac} $$ + $$\sqrt {2ab} $$

Dividing by $$\sqrt {abc} $$ we get.

$$ \Rightarrow $$  $${1 \over {\sqrt a }}$$ = $${1 \over {\sqrt b }}$$ + $${1 \over {\sqrt c }}$$
4
MCQ (Single Correct Answer)

JEE Main 2019 (Online) 9th January Evening Slot

If the circles

x2 + y2 $$-$$ 16x $$-$$ 20y + 164 = r2  

and  (x $$-$$ 4)2 + (y $$-$$ 7)2 = 36

intersect at two distinct points, then :
A
r > 11
B
0 < r < 1
C
r = 11
D
1 < r < 11

Explanation

Circles are x2 + y2 $$-$$ 16x $$-$$ 20y + 164 = r2 $$ \Rightarrow $$ c1 (8, 10)

and (x $$-$$ 4)2 + (y $$-$$ 7)2 = 36

they intersect at two distinct points

$$\left| {{r_1} - {r_2}} \right| < {c_1}{c_2} < {r_1} + {r_2}\left\{ {{c_1}{c_2} = \sqrt {16 + 9} = 5} \right\}$$

Now  $$\left| {r - 6} \right| < 5 < r + 6$$

$$\left| {r - 6} \right| < 5$$

$$ \Rightarrow $$  $$ - 5 < r - 6 < 5$$

$$ \Rightarrow $$  $$1 < r < 11\,\,\,\,\,\,\,\,\,...(i)$$

$$5 < r + 6$$

$$ - 1 < r\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$$

from (i) and (ii)

r $$ \in $$ (1, 11)

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