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Circle

JEE Mathematics · 133 questions · Page 3 of 14 · Click an option or "Show Solution" to reveal answer

Q21
If the lines 3x−4y−7=03x - 4y - 7 = 0 and 2x−3y−5=02x - 3y - 5 = 0 are two diameters of a circle of area 49π49\pi square units, the equation of the circle is :
A  x2+y2+2x −2y−47=0 \,{x^2} + {y^2} + 2x\, - 2y - 47 = 0\,
B  x2+y2+2x −2y−62=0 \,{x^2} + {y^2} + 2x\, - 2y - 62 = 0\,
C x2+y2−2x +2y−62=0{x^2} + {y^2} - 2x\, + 2y - 62 = 0
D x2+y2−2x +2y−47=0{x^2} + {y^2} - 2x\, + 2y - 47 = 0
Correct Answer
Option D
Solution

Point of intersection of

3x−4y−7=03x - 4y - 7 = 0

and

2x−3y−5=02x - 3y - 5 = 0

is

(1,−1)\left( {1, - 1} \right)

which is the center of the circle and radius

=7=7

∴\therefore Equation is

(x−1)2+(y+1)2=49{\left( {x - 1} \right)^2} + {\left( {y + 1} \right)^2} = 49
⇒x2+y2−2x+2y−47=0\Rightarrow {x^2} + {y^2} - 2x + 2y - 47 = 0
Q22
Let CC be the circle with centre (0,0)(0, 0) and radius 33 units. The equation of the locus of the mid points of the chords of the circle CC that subtend an angle of 2π3{{2\pi } \over 3} at its center is :
A x2+y2=32{x^2} + {y^2} = {3 \over 2}
B x2+y2=1{x^2} + {y^2} = 1
C x2+y2=274{x^2} + {y^2} = {{27} \over 4}
D x2+y2=94{x^2} + {y^2} = {{9} \over 4}
Correct Answer
Option D
Solution

Let

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

be the mid point of chord

ABAB

where

∠AOB=2π3\angle AOB = {{2\pi } \over 3}

∴\therefore

∠AOM=π3.\angle AOM = {\pi \over 3}.

Also

OM=OM=
3cos⁡π3=323\cos {\pi \over 3} = {3 \over 2}
⇒h2+k2=32\Rightarrow \sqrt {{h^2} + k{}^2} = {3 \over 2}
⇒h2+k2=94\Rightarrow {h^2} + {k^2} = {9 \over 4}

∴\therefore Locus of

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

is

x2+y2=94{x^2} + {y^2} = {9 \over 4}
Q23
Consider a family of circles which are passing through the point (−1,1)(-1, 1) and are tangent to xx-axis. If (h,k)(h, k) are the coordinate of the centre of the circles, then the set of values of kk is given by the interval :
A −12≤k≤12 - {1 \over 2} \le k \le {1 \over 2}
B k≤12k \le {1 \over 2}
C 0≤k≤120 \le k \le {1 \over 2}
D k≥12k \ge {1 \over 2}
Correct Answer
Option D
Solution

Equation of circle whose center is

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

i.e

(x−h)2+(y−k)2=k2{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {k^2}

(radius of circle

=k=k

because circle is tangent to

xx

-axis) Equation of circle passing through

(−1,+1)\left( { - 1, + 1} \right)

∴\therefore

(−1,−h)2+(1−k)2=k2{\left( { - 1, - h} \right)^2} + {\left( {1 - k} \right)^2} = {k^2}
⇒1+h2+2h+1+k2−2k=k2\Rightarrow 1 + {h^2} + 2h + 1 + {k^2} - 2k = {k^2}
⇒h2+2h−2k+2=0\Rightarrow {h^2} + 2h - 2k + 2 = 0
D≥0D \ge 0

∴\therefore

(2)2−4×1.(−2k+2)≥0{\left( 2 \right)^2} - 4 \times 1.\left( { - 2k + 2} \right) \ge 0
⇒4−4(−2k+2)≥0\Rightarrow 4 - 4\left( { - 2k + 2} \right) \ge 0
⇒1+2k−2≥0\Rightarrow 1 + 2k - 2 \ge 0
⇒k≥12\Rightarrow k \ge {1 \over 2}
Q24
The point diametrically opposite to the point P(1,0)P(1, 0) on the circle x2+y2+2x+4y−3=0{x^2} + {y^2} + 2x + 4y - 3 = 0 is :
A (3,−4)(3, -4)
B (−3,4)(-3, 4)
C (−3,−4)(-3, -4)
D (3,4)(3, 4)
Correct Answer
Option C
Solution

The given circle is

x2+y2+2x+4y−3=0{x^2} + {y^2} + 2x + 4y - 3 = 0

Center

(−1,−2)(-1,-2)

Let

QQ
(α,β)\left( {\alpha ,\beta } \right)

be the point diametrically opposite to the point

P(1,0),P(1,0),

then

1+α2=−1{{1 + \alpha } \over 2} = - 1

and

0+β2=−2{{0 + \beta } \over 2} = - 2
⇒α=−3,β=−4,\Rightarrow \alpha = - 3,\beta = - 4,

So,

QQ

is

(−3,−4)\left( { - 3, - 4} \right)
Q25
The differential equation of the family of circles with fixed radius 55 units and centre on the line y=2y = 2 is :
A (x−2)y2=25−(y−2)2\left( {x - 2} \right){y^2} = 25 - {\left( {y - 2} \right)^2}
B (y−2)y2=25−(y−2)2\left( {y - 2} \right){y^2} = 25 - {\left( {y - 2} \right)^2}
C (y−2)2y2=25−(y−2)2{\left( {y - 2} \right)^2}{y^2} = 25 - {\left( {y - 2} \right)^2}
D (x−2)2y2=25−(y−2)2{\left( {x - 2} \right)^2}{y^2} = 25 - {\left( {y - 2} \right)^2}
Correct Answer
Option C
Solution

Let the center of the circle be

(h,2)(h, 2)

∴\therefore Equation of circle is

(x−h)2+(y−2)2=25         ...(1){\left( {x - h} \right)^2} + \left( {y - 2} \right){}^2 = 25\,\,\,\,\,\,\,\,\,...\left( 1 \right)

Differentiating with respect to

x,x,

we get

2(x−h)+2(y−2)dydx=02\left( {x - h} \right) + 2\left( {y - 2} \right){{dy} \over {dx}} = 0
⇒x−h=−(y−2)dydx\Rightarrow x - h = - \left( {y - 2} \right){{dy} \over {dx}}

Substituting in equation

(1)(1)

we get

(y−2)2(dydx)2+(y−2)2=25{\left( {y - 2} \right)^2}{\left( {{{dy} \over {dx}}} \right)^2} + {\left( {y - 2} \right)^2} = 25
⇒(y−2)2(y′)2=25−(y−2)2\Rightarrow {\left( {y - 2} \right)^2}{\left( {y'} \right)^2} = 25 - {\left( {y - 2} \right)^2}
Q26
Three distinct points A, B and C are given in the 2 -dimensional coordinates plane such that the ratio of the distance of any one of them from the point (1,0)(1, 0) to the distance from the point (−1,0)(-1, 0) is equal to 13{1 \over 3}. Then the circumcentre of the triangle ABC is at the point :
A (54,0)\left( {{5 \over 4},0} \right)
B (52,0)\left( {{5 \over 2},0} \right)
C (53,0)\left( {{5 \over 3},0} \right)
D (0,0)\left( {0,0} \right)
Correct Answer
Option A
Solution

Given that

P(1,0),Q(−1,0)P\left( {1,0} \right),Q\left( { - 1,0} \right)

and

APAQ=BPBQ=CPCQ=13{{AP} \over {AQ}} = {{BP} \over {BQ}} = {{CP} \over {CQ}} = {1 \over 3}
⇒3AP=AQ\Rightarrow 3AP = AQ
      \,\,\,\,\,\,

Let

A=(x,y)A = (x,y)

then

3AP=AQ⇒9AP2=AQ23AP = AQ \Rightarrow 9A{P^2} = A{Q^2}
⇒9(x−1)2+9y2=(x+1)2+y2\Rightarrow 9{\left( {x - 1} \right)^2} + 9{y^2} = {\left( {x + 1} \right)^2} + y{}^2
⇒9x2−18x+9+9y2=x2+2x+1+y2\Rightarrow 9{x^2} - 18x + 9 + 9{y^2} = {x^2} + 2x + 1 + {y^2}
⇒8x2−20x+8y2+8=0\Rightarrow 8{x^2} - 20x + 8{y^2} + 8 = 0
⇒x2+y2−53x+1=0             ...(1)\Rightarrow {x^2} + {y^2} - {5 \over 3}x + 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

∴\therefore A lies on the circle given by eq.

(1).(1).

As

BB

and

CC

also follow the same condition, - they must lie on the same circle. ∴\therefore Center of circumcircle of

ΔABC\Delta ABC

== Center of circle given by

(1)=(54,0)\left( 1 \right) = \left( {{5 \over 4},0} \right)
Q27
If PP and QQ are the points of intersection of the circles x2+y2+3x+7y+2p−5=0{x^2} + {y^2} + 3x + 7y + 2p - 5 = 0 and x2+y2+2x+2y−p2=0{x^2} + {y^2} + 2x + 2y - {p^2} = 0 then there is a circle passing through P,QP,Q and (1,1)(1, 1) for :
A all except one value of pp
B all except two values of pp
C exactly one value of pp
D all values of pp
Correct Answer
Option A
Solution

The given circles are

S1≡x2+y2+3x+7y+2p−5=0          ...(1){S_1} \equiv {x^2} + {y^2} + 3x + 7y + 2p - 5 = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)
S2≡x2+y2+2x+2y−p2=0                ...(2){S_2} \equiv {x^2} + {y^2} + 2x + 2y - {p^2} = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)

∴\therefore Equation of common chord

PQPQ

is

S1−S2=0{S_1} - {S_2} = 0
⇒L≡x+5y+p2+2p−5=0\Rightarrow L \equiv x + 5y + {p^2} + 2p - 5 = 0

⇒\Rightarrow Equation of circle passing through

PP

and

QQ

is

S1+λ  L=0{S_1} + \lambda \,\,L = 0
⇒(x2+y2+3x+7y+2p−5)+λ\Rightarrow \left( {{x^2} + {y^2} + 3x + 7y + 2p - 5} \right) + \lambda
(x+5y+p2+2p−5)=0\left( {x + 5y + {p^2} + 2p - 5} \right) = 0

As it passes through

(1,1),\left( {1,1} \right),

therefore

⇒(7+2p)+λ(2p+p2+1)=0\Rightarrow \left( {7 + 2p} \right) + \lambda \left( {2p + {p^2} + 1} \right) = 0
⇒λ=−2p+7(p+1),\Rightarrow \lambda = - {{2p + 7} \over {\left( {p + 1} \right)}},

which does not exist for

p=−1p=-1
Q28
The circle x2+y2=4x+8y+5{x^2} + {y^2} = 4x + 8y + 5 intersects the line 3x−4y=m3x - 4y = m at two distinct points if :
A −35<m<15 - 35 < m < 15
B 15<m<65 15 < m < 65
C 35<m<85 35 < m < 85
D −85<m<−35 - 85 < m < -35
Correct Answer
Option A
Solution

Circle

x2+y2−4x−8y−5=0{x^2} + {y^2} - 4x - 8y - 5 = 0

Center

=(2,4),=(2,4),

Radius

=4+16+5=5= \sqrt {4 + 16 + 5} = 5

If circle is intersecting line

3x−4y=m,3x-4y=m,

at two distinct points. ⇒\Rightarrow length of perpendicular from center to the line

<<

radius

⇒∣6−16−m∣5<5⇒∣10+m∣<25\Rightarrow {{\left| {6 - 16 - m} \right|} \over 5} < 5 \Rightarrow \left| {10 + m} \right| < 25
⇒−25<m+10<25⇒−35<m<15\Rightarrow - 25 < m + 10 < 25 \Rightarrow - 35 < m < 15
Q29
The two circles x2 + y2 = ax, and x2 + y2 = c2 (c > 0) touch each other if :
A | a | = c
B a = 2c
C | a | = 2c
D 2 | a | = c
Correct Answer
Option A
Solution

As center of one circle is

(0,0)\left( {0,0} \right)

and other circle passes through

(0,0),(0,0),

therefore Also

C1(a2,0)C2(0,0){C_1}\left( {{a \over 2},0} \right){C_2}\left( {0,0} \right)
r1=a2r2=C{r_1} = {a \over 2}{r_2} = C
C1C2=r1−r2=a2{C_1}{C_2} = {r_1} - {r_2} = {a \over 2}
⇒C−a2=a2\Rightarrow C - {a \over 2} = {a \over 2}
⇒C=a\Rightarrow C = a

If the two circles touch each other, then they must touch each other internally.

Q30
The length of the diameter of the circle which touches the xx-axis at the point (1,0)(1, 0) and passes through the point (2,3)(2, 3) is :
A 103{{10} \over 3}
B 35{{3} \over 5}
C 65{{6} \over 5}
D 53{{5} \over 3}
Correct Answer
Option A
Solution

Let center of the circle be

(1,h)\left( {1,h} \right)
[  \left[ {\,\,} \right.

as circle touches

xx

-axis at

(1,0)  ]\left. {\left( {1,0} \right)\,\,} \right]

Let the circle passes through the point

B(2,3)B(2,3)

∴\therefore

CA=CBCA=CB

(radius)

⇒CA2=CB2\Rightarrow C{A^2} = C{B^2}
⇒(1−1)2+(h−0)2=(1−2)2+(h−3)2\Rightarrow {\left( {1 - 1} \right)^2} + \left( {h - 0} \right){}^2 = {\left( {1 - 2} \right)^2} + {\left( {h - 3} \right)^2}
⇒h2=1+h2+9−6h\Rightarrow {h^2} = 1 + {h^2} + 9 - 6h
⇒h=106=53\Rightarrow h = {{10} \over 6} = {5 \over 3}

Thus, diameter is

2h=103.2h = {{10} \over 3}.
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