Practice Start Test

Circle

JEE Mathematics · 133 questions · Page 2 of 14 · Click an option or "Show Solution" to reveal answer

Q11
The centres of a set of circles, each of radius 3, lie on the circle x2 + y2=25{x^2}\, + \,{y^2} = 25. The locus of any point in the set is :
A 4 ≤  x2 + y2 ≤  644\, \le \,\,{x^2}\, + \,{y^2}\, \le \,\,64
B x2 + y2 ≤  25{x^2}\, + \,{y^2}\, \le \,\,25
C x2 + y2 ≥  25{x^2}\, + \,{y^2}\, \ge \,\,25
D 3 ≤  x2 + y2 ≤  93\, \le \,\,{x^2}\, + \,{y^2}\, \le \,\,9
Correct Answer
Option A
Solution

For any point

P(x,y)P(x,y)

in the given circle, we should have

OA≤OP≤OBOA \le OP \le OB
⇒(5−3)≤x2+y2≤5+3\Rightarrow \left( {5 - 3} \right) \le \sqrt {{x^2} + {y^2}} \le 5 + 3
⇒4≤x2+y2≤64\Rightarrow 4 \le {x^2} + {y^2} \le 64
Q12
If the two circles (x−1)2 + (y−3)2= r2{(x - 1)^2}\, + \,{(y - 3)^2} = \,{r^2} and  x2 + y2− 8x + 2y +  8  =0\,{x^2}\, + \,{y^2} - \,8x\, + \,2y\, + \,\,8\,\, = 0 intersect in two distinct point, then :
A r>2r > 2
B 2<r<82 < r < 8
C r<2r < 2
D r=2.r = 2.
Correct Answer
Option B
Solution
∣r1−r2∣<C1C2\left| {{r_1} - {r_2}} \right| < {C_1}{C_2}

for intersection

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

and

r1+r2>C1C2, {r_1} + {r_2} > {C_1}{C_2},\,
r+3>5⇒r>2   ...(2)r + 3 > 5 \Rightarrow r > 2\,\,\,...\left( 2 \right)

From

(1)\left( 1 \right)

and

(2),\left( 2 \right),
2<r<8.2 < r < 8.
Q13
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 x2 + y2− 2x + 2y  = 62{x^2}\, + \,{y^2} - \,2x\, + \,2y\,\, = \,62
B x2 + y2+ 2x − 2y  = 62{x^2}\, + \,{y^2} + \,2x\, - \,2y\,\, = \,62
C x2 + y2+ 2x − 2y  = 47{x^2}\, + \,{y^2} + \,2x\, - \,2y\,\, = \,47
D x2 + y2− 2x + 2y  = 47{x^2}\, + \,{y^2} - \,2x\, + \,2y\,\, = \,47
Correct Answer
Option D
Solution
πr2=154⇒r=7\pi {r^2} = 154 \Rightarrow r = 7

For center on solving equation

2x−3y=5&3x−4y=72x - 3y = 5\& 3x - 4y = 7

we get

x=1, y=−1x = 1,\,y = - 1

∴\therefore center

=(1,−1)=(1,-1)

Equation of circle,

(x−1)2+(y+1)2=72{\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {7^2}
x2+y2−2x+2y=47{x^2} + {y^2} - 2x + 2y = 47
Q14
A variable circle passes through the fixed point A (p, q) and touches x-axis. The locus of the other end of the diameter through A is :
A (y − q)2= 4 px{(y\, - \,q)^2} = \,4\,px
B (x − q)2= 4 py{(x\, - \,q)^2} = \,4\,py
C (y − p)2= 4 qx{(y\, - \,p)^2} = \,4\,qx
D (x − p)2= 4 qy{(x\, - \,p)^2} = \,4\,qy
Correct Answer
Option D
Solution

Let the variable circle be

x2+y2+2gx+2fy+c=0          ...(1){x^2} + {y^2} + 2gx + 2fy + c = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

∴\therefore

p2+q2+2gp+2fq+c=0         ...(2){p^2} + {q^2} + 2gp + 2fq + c = 0\,\,\,\,\,\,\,\,\,...\left( 2 \right)

Circle

(1)(1)

touches

xx

-axis, ∴\therefore

g2−c=0⇒c=g2.   {g^2} - c = 0 \Rightarrow c = {g^2}.\,\,\,

From

(2)(2)
p2+q2+2gp+2fq+g2=0         ...(3){p^2} + {q^2} + 2gp + 2fq + {g^2} = 0\,\,\,\,\,\,\,\,\,...\left( 3 \right)

Let the other end of diameter through

(p,q)(p,q)

be

(h,k),(h,k),

then,

h+p2=−g{{h + p} \over 2} = - g

and

k+q2=−f{{k + q} \over 2} = - f

Put in

(3)(3)
p2+q2+2p(−h+p2)+2q(−k+q2)+(h+p2)2=0{p^2} + {q^2} + 2p\left( { - {{h + p} \over 2}} \right) + 2q\left( { - {{k + q} \over 2}} \right) + {\left( {{{h + p} \over 2}} \right)^2} = 0
⇒h2+p2−2hp−4kq=0\Rightarrow {h^2} + {p^2} - 2hp - 4kq = 0

∴\therefore locus of

(h,k)\left( {h,k} \right)

is

x2+p2−2xp−4yq=0{x^2} + {p^2} - 2xp - 4yq = 0
⇒(x−p)2=4qy\Rightarrow {\left( {x - p} \right)^2} = 4qy
Q15
If the lines 2x + 3y + 1 + 0 and 3x - y - 4 = 0 lie along diameter of a circle of circumference 10 π10\,\pi , then the equation of the circle is :
A x2 + y2+ 2x − 2y− 23  =0{x^2}\, + \,{y^2} + \,2x\, - \,2y - \,23\,\, = 0
B x2 + y2− 2x − 2y− 23  =0{x^2}\, + \,{y^2} - \,2x\, - \,2y - \,23\,\, = 0
C x2 + y2+ 2x + 2y− 23  =0{x^2}\, + \,{y^2} + \,2x\, + \,2y - \,23\,\, = 0
D x2 + y2− 2x + 2y− 23  =0{x^2}\, + \,{y^2} - \,2x\, + \,2y - \,23\,\, = 0
Correct Answer
Option D
Solution

Two diameters are along

2x+3y+1=02x+3y+1=0

and

3x−y−4=03x-y-4=0

solving we get center

(1,−1)(1,-1)

circumference

=2πr=10π= 2\pi r = 10\pi

∴\therefore

r=5r=5

. Required circle is,

(x−1)2+(y+1)2=52{\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = {5^2}
⇒x2+y2−2x+2y−23=0\Rightarrow {x^2} + {y^2} - 2x + 2y - 23 = 0
Q16
If a circle passes through the point (a, b) and cuts the circle x2 + y2=4{x^2}\, + \,{y^2} = 4 orthogonally, then the locus of its centre is :
A 2ax −2by −(a2 + b2+4)=02ax\, - 2by\, - ({a^2}\, + \,{b^2} + 4) = 0
B 2ax +2by −(a2 + b2+4)=02ax\, + 2by\, - ({a^2}\, + \,{b^2} + 4) = 0
C 2ax −2by +(a2 + b2+4)=02ax\, - 2by\, + ({a^2}\, + \,{b^2} + 4) = 0
D 2ax +2by +(a2 + b2+4)=02ax\, + 2by\, + ({a^2}\, + \,{b^2} + 4) = 0
Correct Answer
Option B
Solution

Let the variable circle is

x2+y2+2gx+2fy+c=0        ...(1){x^2} + {y^2} + 2gx + 2fy + c = 0\,\,\,\,\,\,\,\,...\left( 1 \right)

It passes through

(a,b)(a,b)

∴\therefore

a2+b2+2ga+2fb+c=0       ...(2){a^2} + {b^2} + 2ga + 2fb + c = 0\,\,\,\,\,\,\,...\left( 2 \right)
(1)(1)

cuts

x2+y2=4{x^2} + {y^2} = 4

orthogonally ∴\therefore

2(g×0+f×0)=c−4⇒c=42\left( {g \times 0 + f \times 0} \right) = c - 4 \Rightarrow c = 4

∴\therefore from

(2)(2)
   a2+b2+2ga+2fb+4=0\,\,\,{a^2} + {b^2} + 2ga + 2fb + 4 = 0

∴\therefore Locus of center

(−g,−f)\left( { - g, - f} \right)

is

a2+b2−2ax−2by+4=0{a^2} + {b^2} - 2ax - 2by + 4 = 0

or

2ax+2by=a2+b2+42ax + 2by = {a^2} + {b^2} + 4
Q17
A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius 2. The locus of the centre of the circle is :
A an ellipse
B a circle
C a hyperbola
D a parabola
Correct Answer
Option D
Solution

Equation of circle with center

(0,3)(0,3)

and radius

22

is

x2+(y−3)2=4{x^2} + {\left( {y - 3} \right)^2} = 4

Let locus of the variable circle is

(α,β)\left( {\alpha ,\beta } \right)

As it touches

xx

-axis. ∴\therefore It's equation is

(x−α)2+(y+β)2=β2{\left( {x - \alpha } \right)^2} + {\left( {y + \beta } \right)^2} = {\beta ^2}

Circle touch externally

⇒c1c2=r1+r2\Rightarrow {c_1}{c_2} = {r_1} + {r_2}

∴\therefore

α2+(β−3)2=2+β\sqrt {{\alpha ^2} + {{\left( {\beta - 3} \right)}^2}} = 2 + \beta
α2+(β−3)2=β2+4+4β{\alpha ^2} + {\left( {\beta - 3} \right)^2} = {\beta ^2} + 4 + 4\beta
⇒α2=10(β−1/2)\Rightarrow {\alpha ^2} = 10\left( {\beta - 1/2} \right)

∴\therefore Locus is

x2=10(y−12){x^2} = 10\left( {y - {1 \over 2}} \right)

which is parabola.

Q18
If the pair of lines ax2+2(a+b)xy+by2=0a{x^2} + 2\left( {a + b} \right)xy + b{y^2} = 0 lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then :
A 3a2−10ab+3b2=03{a^2} - 10ab + 3{b^2} = 0
B 3a2−2ab+3b2=03{a^2} - 2ab + 3{b^2} = 0
C 3a2+10ab+3b2=03{a^2} + 10ab + 3{b^2} = 0
D 3a2+2ab+3b2=03{a^2} + 2ab + 3{b^2} = 0
Correct Answer
Option D
Solution

As per question area of one sector

=3=3

area of another sector ⇒\Rightarrow at center by one sector

=3×= 3 \times

angle at center by another sector Let one angle be θ\theta then other

=30=30

Clearly

θ+3θ=180⇒θ=45∘\theta + 3\theta = 180 \Rightarrow \theta = {45^ \circ }

∴\therefore Angle between the diameters represented by combined equation

ax2+2(a+b   xy)+by2=0a{x^2} + 2\left( {a + b\,\,\,xy} \right) + b{y^2} = 0

is

45∘{45^ \circ }

∴\therefore Using

tantan

θ\theta

=2h2−aba+b= {{2\sqrt {{h^2} - ab} } \over {a + b}}

we get

tan⁡ 45∘=2(a+b)2−aba+b\tan \,{45^ \circ } = {{2\sqrt {{{\left( {a + b} \right)}^2} - ab} } \over {a + b}}
⇒1=2a2+b2+aba+b\Rightarrow 1 = {{2\sqrt {{a^2} + {b^2} + ab} } \over {a + b}}
⇒(a+b)2=4(a2+b2+ab)\Rightarrow {\left( {a + b} \right)^2} = 4\left( {{a^2} + {b^2} + ab} \right)
⇒a2+b2+2ab=4a2+4b2+4ab\Rightarrow {a^2} + {b^2} + 2ab = 4{a^2} + 4{b^2} + 4ab
⇒3a2+3b2+2ab=0\Rightarrow 3{a^2} + 3{b^2} + 2ab = 0
Q19
If the circles x2 + y2+ 2ax + cy +a  =0{x^2}\, + \,{y^2} + \,2ax\, + \,cy\, + a\,\, = 0 and x2 + y2− 3ax + dy −1  =0{x^2}\, + \,{y^2} - \,3ax\, + \,dy\, - 1\,\, = 0 intersect in two ditinct points P and Q then the line 5x + by - a = 0 passes through P and Q for :
A exactly one value of a
B no value of a
C infinitely many values of a
D exactly two values of a
Correct Answer
Option B
Solution
s1=x2+y2+2ax+cy+a=0{s_1} = {x^2} + {y^2} + 2ax + cy + a = 0
s2=x2+y2−3ax+dy−1=0{s_2} = {x^2} + {y^2} - 3ax + dy - 1 = 0

Equation of common chord of circles

s1{s_1}

and

s2{s_2}

is given by

s1−s2=0{s_1} - {s_2} = 0
⇒5ax+(c−d)y+a+1=0\Rightarrow 5ax + \left( {c - d} \right)y + a + 1 = 0

Given that

5x+by−a=05x + by - a = 0

passes through

PP

and

QQ

∴\therefore The two equations should represent the same line

⇒a1=c−db=a+1−a\Rightarrow {a \over 1} = {{c - d} \over b} = {{a + 1} \over { - a}}
⇒a+1=−a2\Rightarrow a + 1 = - {a^2}
a2+a+1=0{a^2} + a + 1 = 0

No real value of

a.a.
Q20
If a circle passes through the point (a, b) and cuts the circle x2 + y2=p2{x^2}\, + \,{y^2} = {p^2} orthogonally, then the equation of the locus of its centre is :
A x2 + y2− 3ax − 4 by  + (a2 + b2−p2)=0{x^2}\, + \,{y^2} - \,3ax\, - \,4\,by\,\, + \,({a^2}\, + \,{b^2} - {p^2}) = 0
B 2ax +  2 by  − (a2 − b2+p2)=02ax\, + \,\,2\,by\,\, - \,({a^2}\, - \,{b^2} + {p^2}) = 0
C x2 + y2− 2ax −  3 by  + (a2 − b2−p2)=0{x^2}\, + \,{y^2} - \,2ax\, - \,\,3\,by\,\, + \,({a^2}\, - \,{b^2} - {p^2}) = 0
D 2ax +  2 by  − (a2 + b2+p2)=02ax\, + \,\,2\,by\,\, - \,({a^2}\, + \,{b^2} + {p^2}) = 0
Correct Answer
Option D
Solution

Let the center be

(α,β)\left( {\alpha ,\beta } \right)

As It cuts the circle

x2+y2=p2{x^2} + {y^2} = {p^2}

orthogonally ∴\therefore Using

2g1g2+2f1f2=c1+c2,  2{g_1}{g_2} + 2{f_1}{f_2} = {c_1} + {c_2},\,\,

we get

2(−α)×0+2(−β)×02\left( { - \alpha } \right) \times 0 + 2\left( { - \beta } \right) \times 0
=c1−p2⇒c1=p2= {c_1} - {p^2} \Rightarrow {c_1} = {p^2}

Let equation of circle is

x2+y2−2αx−2βy+p2=0{x^2} + {y^2} - 2\alpha x - 2\beta y + {p^2} = 0

It passes through

(a,b)⇒a2+b2−2αa−2βb+p2=0\left( {a,b} \right) \Rightarrow {a^2} + {b^2} - 2\alpha a - 2\beta b + {p^2} = 0

∴\therefore Locus of

(α,β)\left( {\alpha ,\beta } \right)

is ∴\therefore

2ax+2by−(a2+b2+p2)=0.2ax + 2by - \left( {{a^2} + {b^2} + {p^2}} \right) = 0.
Ready for a full JEE mock test? Timed · full syllabus · instant results
Take a Mock Test →