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Circle

JEE Mathematics · 133 questions · Page 9 of 14 · Click an option or "Show Solution" to reveal answer

Q81
Let a circle C touch the lines L1:4x−3y+K1=0{L_1}:4x - 3y + {K_1} = 0 and L2=4x−3y+K2=0{L_2} = 4x - 3y + {K_2} = 0, K1,K2∈R{K_1},{K_2} \in R. If a line passing through the centre of the circle C intersects L1 at (−1,2)( - 1,2) and L2 at (3,−6)(3, - 6), then the equation of the circle C is :
A (x−1)2+(y−2)2=4{(x - 1)^2} + {(y - 2)^2} = 4
B (x+1)2+(y−2)2=4{(x + 1)^2} + {(y - 2)^2} = 4
C (x−1)2+(y+2)2=16{(x - 1)^2} + {(y + 2)^2} = 16
D (x−1)2+(y−2)2=16{(x - 1)^2} + {(y - 2)^2} = 16
Correct Answer
Option C
Solution

Co-ordinate of centre

C≡(3+(−1)2,−6−22)≡(1,−2)C \equiv \left( {{{3 + ( - 1)} \over 2},{{ - 6 - 2} \over 2}} \right) \equiv (1, - 2)

L1 is passing through A

⇒−4−6+K1=0\Rightarrow - 4 - 6 + {K_1} = 0
⇒K1=10\Rightarrow {K_1} = 10

L2 is passing through B

⇒12+18+K2=0\Rightarrow 12 + 18 + {K_2} = 0
⇒K2=−30\Rightarrow {K_2} = - 30

Equation of

L1:4x−3y+10=0{L_1}:4x - 3y + 10 = 0

Equation of

L2=4x−3y−30=0{L_2} = 4x - 3y - 30 = 0

Diameter of circle

=∣10+3042+(−3)2∣=8= \left| {{{10 + 30} \over {\sqrt {{4^2} + {{( - 3)}^2}} }}} \right| = 8

⇒\Rightarrow Radius = 4 Equation of circle

(x−1)2+(y+2)2=16{(x - 1)^2} + {(y + 2)^2} = 16
Q82
Consider three circles: C1:x2+y2=r2{C_1}:{x^2} + {y^2} = {r^2} C2:(x−1)2+(y−1)2=r2{C_2}:{(x - 1)^2} + {(y - 1)^2} = {r^2} C3:(x−2)2+(y−1)2=r2{C_3}:{(x - 2)^2} + {(y - 1)^2} = {r^2} If a line L : y = mx + c be a common tangent to C1, C2 and C3 such that C1 and C3 lie on one side of line L while C2 lies on other side, then the value of 20(r2+c)20({r^2} + c) is equal to :
A 23
B 15
C 12
D 6
Correct Answer
Option D
Solution
c1:x2+y2=r2{c_1}:{x^2} + {y^2} = {r^2}

; center = (0, 0) and radius = r

c2:(x−1)2+(y−1)2=r2{c_2}:{(x - 1)^2} + {(y - 1)^2} = {r^2}

; center = (1, 1) and radius = r

c3:(x−2)2+(y−1)2=r2{c_3}:{(x - 2)^2} + {(y - 1)^2} = {r^2}

; center = (2, 1) and radius = r Distance of

y=mx+cy = mx + c

line from center (0, 0) is,

∣0+0+cm2+1∣=r\left| {{{0 + 0 + c} \over {\sqrt {{m^2} + 1} }}} \right| = r

..... (1) Distance of

y=mx+cy = mx + c

line from center (1, 1) is,

∣m−1+cm2+1∣=r\left| {{{m - 1 + c} \over {\sqrt {{m^2} + 1} }}} \right| = r

..... (2) Distance of

y=mx+cy = mx + c

line from center (2, 1) is,

∣2m−1+cm2+1∣=r\left| {{{2m - 1 + c} \over {\sqrt {{m^2} + 1} }}} \right| = r

.... (3) From (1) and (2), we get

∣c1+m2∣=∣m−1+c1+m2∣\left| {{c \over {\sqrt {1 + {m^2}} }}} \right| = \left| {{{m - 1 + c} \over {\sqrt {1 + {m^2}} }}} \right|
⇒m−1+c=± c\Rightarrow m - 1 + c = \pm \,c

..... (4) taking positive sign,

m−1+c=cm - 1 + c = c
⇒m−1=0\Rightarrow m - 1 = 0
⇒m=1\Rightarrow m = 1

From (2) and (3), we get

∣m−1+c1+m2∣=∣2m−1+cm2+1∣\left| {{{m - 1 + c} \over {\sqrt {1 + {m^2}} }}} \right| = \left| {{{2m - 1 + c} \over {\sqrt {{m^2} + 1} }}} \right|
⇒(m−1+c)= ± (2m−1+c)\Rightarrow (m - 1 + c) = \, \pm \,(2m - 1 + c)

...... (5) taking positive sign,

m−1+c=2m−1+cm - 1 + c = 2m - 1 + c
⇒m=0\Rightarrow m = 0

By taking positive sign we get two different value of m so it is not acceptable.

From equation (4), taking negative sign,

m−1+c=−cm - 1 + c = - c
⇒m+2c−1=0\Rightarrow m + 2c - 1 = 0

..... (6) From equation (5), taking negative sign

m−1+c=−(2m−1+c)m - 1 + c = - (2m - 1 + c)
⇒3m+2c−2=0\Rightarrow 3m + 2c - 2 = 0

..... (7) Solving equation (6) and (7), we get

3m+1−m−2=03m + 1 - m - 2 = 0
⇒2m=1\Rightarrow 2m = 1
⇒m=12\Rightarrow m = {1 \over 2}

∴\therefore

2c=1−122c = 1 - {1 \over 2}
⇒c=14\Rightarrow c = {1 \over 4}

Putting value of

m=12m = {1 \over 2}

and

c=14c = {1 \over 4}

in equation (1), we get

r=∣141+14∣r = \left| {{{{1 \over 4}} \over {\sqrt {1 + {1 \over 4}} }}} \right|
=∣14×25∣= \left| {{1 \over 4} \times {2 \over {\sqrt 5 }}} \right|
=125= {1 \over {2\sqrt 5 }}

∴\therefore

20(r2+c)20({r^2} + c)
=20(14×5+14)= 20\left( {{1 \over {4 \times 5}} + {1 \over 4}} \right)
=20(1+520)= 20\left( {{{1 + 5} \over {20}}} \right)
=6= 6
Q83
A circle C1C_{1} passes through the origin O\mathrm{O} and has diameter 4 on the positive xx-axis. The line y=2xy=2 x gives a chord OA\mathrm{OA} of circle C1\mathrm{C}_{1}. Let C2\mathrm{C}_{2} be the circle with OA\mathrm{OA} as a diameter. If the tangent to C2\mathrm{C}_{2} at the point A\mathrm{A} meets the xx-axis at P\mathrm{P} and yy-axis at Q\mathrm{Q}, then QA:AP\mathrm{QA}: \mathrm{AP} is equal to :
A 1 : 4
B 1 : 5
C 2 : 5
D 1 : 3
Correct Answer
Option A
Solution

Equation of C1

x2+y2−4x=0{x^2} + {y^2} - 4x = 0

Intersection with

y=2xy = 2x
x2+4x2−4x=0{x^2} + 4{x^2} - 4x = 0
5x2−4x=05{x^2} - 4x = 0
⇒x=0,45\Rightarrow x = 0,{4 \over 5}
y=0,85y = 0,{8 \over 5}
A:(45,85)A:\left( {{4 \over 5},{8 \over 5}} \right)

Tangent of C2 at

A(45,85)A\left( {{4 \over 5},{8 \over 5}} \right)
x+2y=4⇒P:(4,0), Q:(0,2)x + 2y = 4 \Rightarrow P:(4,0),\,Q:(0,2)
QA:AP=1:4QA:AP = 1:4
Q84
For t∈(0,2π)\mathrm{t} \in(0,2 \pi), if ABC\mathrm{ABC} is an equilateral triangle with vertices A(sin⁡t,−cos⁡t),B(cost⁡,sin⁡t)\mathrm{A}(\sin t,-\cos \mathrm{t}), \mathrm{B}(\operatorname{cost}, \sin t) and C(a,b)C(a, b) such that its orthocentre lies on a circle with centre (1,13)\left(1, \dfrac{1}{3}\right), then (a2−b2)\left(a^{2}-b^{2}\right) is equal to :
A 83\dfrac{8}{3}
B 8
C 779\dfrac{77}{9}
D 809\dfrac{80}{9}
Correct Answer
Option B
Solution

Let

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

be the orthocentre of

Δ\Delta

ABC Then

h=sin⁡t+cos⁡t+a3, k=−cos⁡t+sin⁡t+b3h = {{\sin t + \cos t + a} \over 3},\,k = {{ - \cos t + \sin t + b} \over 3}

(Orthocentre coincide with centroid) ∴\therefore

(3h−a)2+(3k−b)2=2{(3h - a)^2} + {(3k - b)^2} = 2

∴\therefore

(h−a3)2+(k−b3)2=29{\left( {h - {a \over 3}} \right)^2} + {\left( {k - {b \over 3}} \right)^2} = {2 \over 9}

∵\because Orthocentre lies on circle with centre

(1,13)\left( {1,{1 \over 3}} \right)

∴\therefore

a=3, b=1a = 3,\,b = 1

∴\therefore

a2−b2=8{a^2} - {b^2} = 8
Q85
Let CC be the centre of the circle x2+y2−x+2y=114x^{2}+y^{2}-x+2 y=\dfrac{11}{4} and PP be a point on the circle. A line passes through the point C\mathrm{C}, makes an angle of π4\dfrac{\pi}{4} with the line CP\mathrm{CP} and intersects the circle at the points QQ and RR. Then the area of the triangle PQRP Q R (in unit 2^{2} ) is :
A 2
B 22\sqrt2
C 8sin⁡(π8)8 \sin \left(\dfrac{\pi}{8}\right)
D 8cos⁡(π8)8 \cos \left(\dfrac{\pi}{8}\right)
Correct Answer
Option B
Solution
QR=2r=4QR = 2r = 4
P=(12+2cos⁡π4,−1+2sin⁡π4)P = \left( {{1 \over 2} + 2\cos {\pi \over 4}, - 1 + 2\sin {\pi \over 4}} \right)
=(12+2,−1+2)= \left( {{1 \over 2} + \sqrt 2 , - 1 + \sqrt 2 } \right)

Area of

ΔPQR=12×4×2\Delta PQR = {1 \over 2} \times 4 \times \sqrt 2
=22= 2\sqrt 2

sq. units

Q86
Let the tangents at two points A\mathrm{A} and B\mathrm{B} on the circle x2+y2−4x+3=0x^{2}+\mathrm{y}^{2}-4 x+3=0 meet at origin O(0,0)\mathrm{O}(0,0). Then the area of the triangle OAB\mathrm{OAB} is :
A 332\dfrac{3 \sqrt{3}}{2}
B 334\dfrac{3 \sqrt{3}}{4}
C 323\dfrac{3}{2 \sqrt{3}}
D 343\dfrac{3}{4 \sqrt{3}}
Correct Answer
Option B
Solution
x2+y2−4x+3=0{x^2} + {y^2} - 4x + 3 = 0
⇒(x−2)2+y2=1\Rightarrow {(x - 2)^2} + {y^2} = 1
AO=(OC)2−(AC)2AO = \sqrt {{{(OC)}^2} - {{(AC)}^2}}
=4−1=3= \sqrt {4 - 1} = \sqrt 3
sin⁡θ=12⇒θ=π6\sin \theta = {1 \over 2} \Rightarrow \theta = {\pi \over 6}

Also,

AO=BOAO = BO

Area of

ΔOAB=12 . OA . OBsin⁡60∘\Delta OAB = {1 \over 2}\,.\,OA\,.\,OB\sin 60^\circ
=12×3 . 3 . 32=334= {1 \over 2} \times \sqrt 3 \,.\,\sqrt 3 \,.\,{{\sqrt 3 } \over 2} = {{3\sqrt 3 } \over 4}
Q87
The set of all values of a2a^{2} for which the line x+y=0x+y=0 bisects two distinct chords drawn from a point P(1+a2,1−a2)\mathrm{P}\left(\dfrac{1+a}{2}, \dfrac{1-a}{2}\right) on the circle 2x2+2y2−(1+a)x−(1−a)y=02 x^{2}+2 y^{2}-(1+a) x-(1-a) y=0, is equal to :
A (0,4](0,4]
B (4,∞)(4, \infty)
C (2,12](2,12]
D (8,∞)(8, \infty)
Correct Answer
Option D
Solution

x2+y2−(1+a)x2−(1−a)y2=0x^{2}+y^{2}-\dfrac{(1+a) x}{2}-\dfrac{(1-a) y}{2}=0 Centre (1+a4,1−a4)⇒(h,k)\left(\dfrac{1+\mathrm{a}}{4}, \dfrac{1-\mathrm{a}}{4}\right) \Rightarrow(\mathrm{h}, \mathrm{k}) P(1+a2,1−a2)⇒(2h,2k)\mathrm{P}\left(\dfrac{1+\mathrm{a}}{2}, \dfrac{1-\mathrm{a}}{2}\right) \Rightarrow(2 \mathrm{h}, 2 \mathrm{k}) Equation of chord ⇒T=S1\Rightarrow \mathrm{T}=\mathrm{S}_{1} ⇒(x−y)λ−2 h(x+λ)2−(2k)(y−λ)2\Rightarrow(\mathrm{x}-\mathrm{y}) \lambda-\dfrac{2 \mathrm{~h}(\mathrm{x}+\lambda)}{2}-\dfrac{(2 \mathrm{k})(\mathrm{y}-\lambda)}{2} =2λ2−2 h(λ)+2kλ=2 \lambda^{2}-2 \mathrm{~h}(\lambda)+2 \mathrm{k} \lambda Now, λ(2 h,2k)\lambda(2 \mathrm{~h}, 2 \mathrm{k}) satisfies the chord ∴(2 h−2k)λ−h(x+λ)−k(y−λ)\therefore(2 \mathrm{~h}-2 \mathrm{k}) \lambda-\mathrm{h}(\mathrm{x}+\lambda)-\mathrm{k}(\mathrm{y}-\lambda) ⇒2λ2+4kλ−4 hλ+hλ−kλ+hx+ky=0\Rightarrow 2 \lambda^{2}+4 \mathrm{k} \lambda-4 \mathrm{~h} \lambda+\mathrm{h} \lambda-\mathrm{k} \lambda+\mathrm{hx}+\mathrm{ky}=0 ⇒2λ2+λ(3k−3 h)+ky+hx=0\Rightarrow 2 \lambda^{2}+\lambda(3 \mathrm{k}-3 \mathrm{~h})+\mathrm{ky}+\mathrm{hx}=0 ⇒D>0\Rightarrow \mathrm{D}>0 ⇒9(k−h)2−8(ky+hx)>0\Rightarrow 9(\mathrm{k}-\mathrm{h})^{2}-8(\mathrm{ky}+\mathrm{hx})>0 ⇒9(k−h)2−8(2k2+2 h2)>0\Rightarrow 9(\mathrm{k}-\mathrm{h})^{2}-8\left(2 \mathrm{k}^{2}+2 \mathrm{~h}^{2}\right)>0 ⇒−7k2−7 h2−18kh>0\Rightarrow-7 \mathrm{k}^{2}-7 \mathrm{~h}^{2}-18 \mathrm{kh}>0 ⇒7k2+7 h2+18kh<0\Rightarrow 7 \mathrm{k}^{2}+7 \mathrm{~h}^{2}+18 \mathrm{kh}<0 ⇒7(1−a4)2+7(1+a4)2+18(1−a216)<0\Rightarrow 7\left(\dfrac{1-\mathrm{a}}{4}\right)^{2}+7\left(\dfrac{1+\mathrm{a}}{4}\right)^{2}+18\left(\dfrac{1-\mathrm{a}^{2}}{16}\right)<0 ⇒7[2(1+a2)16]+18(1−a2)16<0,a2=t\Rightarrow 7\left[\dfrac{2\left(1+\mathrm{a}^{2}\right)}{16}\right]+\dfrac{18\left(1-\mathrm{a}^{2}\right)}{16}<0, \quad \mathrm{a}^{2}=\mathrm{t} ⇒78(1+t)+18(1−t)16<0\Rightarrow \dfrac{7}{8}(1+\mathrm{t})+\dfrac{18(1-\mathrm{t})}{16}<0 ⇒14+14t+18−18t16<0\Rightarrow \dfrac{14+14 \mathrm{t}+18-18 \mathrm{t}}{16}<0 ⇒4t>32\Rightarrow 4 \mathrm{t}>32 ⇒\Rightarrow t>8\mathrm{t}>8 ⇒\Rightarrow a2>8 \mathrm{a}^{2}>8

Q88
Four distinct points (2k,3k),(1,0),(0,1)(2 k, 3 k),(1,0),(0,1) and (0,0)(0,0) lie on a circle for kk equal to :
A 313\dfrac{3}{13}
B 213\dfrac{2}{13}
C 513\dfrac{5}{13}
D 113\dfrac{1}{13}
Correct Answer
Option C
Solution
(2k,3k)(2 k, 3 k)

will lie on circle whose diameter is

ABA B

.

(x−1)(x)+(y−1)(y)=0x2+y2−x−y=0.... (i)\begin{aligned} & (x-1)(x)+(y-1)(y)=0 \\ & x^2+y^2-x-y=0 \quad \text{.... (i)} \end{aligned}

Satisfy (2k, 3k) in (i)

(2k)2+(3k)2−2k−3k=013k2−5k=0k=0,k=513 hence k=513\begin{aligned} & (2 k)^2+(3 k)^2-2 k-3 k=0 \\ & 13 k^2-5 k=0 \\ & k=0, k=\frac{5}{13} \\ & \text{ hence } k=\frac{5}{13} \end{aligned}
Q89
Let a circle C1C_{1} be obtained on rolling the circle x2+y2−4x−6y+11=0x^{2}+y^{2}-4 x-6 y+11=0 upwards 4 units on the tangent T\mathrm{T} to it at the point (3,2)(3,2). Let C2C_{2} be the image of C1C_{1} in T\mathrm{T}. Let AA and BB be the centers of circles C1C_{1} and C2C_{2} respectively, and MM and NN be respectively the feet of perpendiculars drawn from AA and BB on the xx-axis. Then the area of the trapezium AMNB is :
A 2(2+2)2\left( {2 + \sqrt 2 } \right)
B 4(1+2)4\left( {1 + \sqrt 2 } \right)
C 3+223 + 2\sqrt 2
D 2(1+2)2\left( {1 + \sqrt 2 } \right)
Correct Answer
Option B
Solution

Given circle is x2+y2−4x−6y+11=0x^{2}+y^{2}-4 x-6 y+11=0, centre (2,3)(2,3) Tangent at (3,2)(3,2) is x−y=1x-y=1 After rolling up by 4 units centre of C1C_{1} is A≡(2+42,3+42)A \equiv\left(2+\dfrac{4}{\sqrt{2}}, 3+\dfrac{4}{\sqrt{2}}\right) ⇒A=(2+22,3+22)\Rightarrow A=(2+2 \sqrt{2}, 3+2 \sqrt{2}) BB is the image of AA in x−y=1x-y=1 x−(2+22)1=y−(3+22)−1=−2(−2)2=2\dfrac{x-(2+2 \sqrt{2})}{1}=\dfrac{y-(3+2 \sqrt{2})}{-1}=\dfrac{-2(-2)}{2}=2 ⇒x=4+22,y=1+22\Rightarrow x=4+2 \sqrt{2}, y=1+2 \sqrt{2} Area of AMNBA M N B =12(4+42)(4+22−(2+22))=\dfrac{1}{2}(4+4 \sqrt{2})(4+2 \sqrt{2}-(2+2 \sqrt{2})) =4(1+2)=4(1+\sqrt{2})

Q90
Let y=x+2,4y=3x+6y=x+2,4y=3x+6 and 3y=4x+13y=4x+1 be three tangent lines to the circle (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2. Then h+kh+k is equal to :
A 6
B 5 (1 + 2\sqrt2)
C 5
D 52\sqrt2
Correct Answer
Option C
Solution
(h,k)=(5.5+5(−2)+14210+72,35+21210+72)(h,k) = \left( {{{5.5 + 5( - 2) + 14\sqrt 2 } \over {10 + 7\sqrt 2 }},{{35 + 21\sqrt 2 } \over {10 + 7\sqrt 2 }}} )\right.
h+k=50+35210+72=5h + k = {{50 + 35\sqrt 2 } \over {10 + 7\sqrt 2 }} = 5
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