1 Crore+ students have signed up on EduRev. Have you? 
If the chord y = mx + 1 of the circle x^{2}+y^{2}=1 subtends an angle of measure 45° at the major segment of the circle then value of m is [2002]
Equation of circle x^{2} + y^{2} = 1 = (1)^{2 }
⇒ x^{2} + y^{2} = (y – mx)^{2} ⇒ x^{2} = m^{2}x^{2} – 2 mxy;
⇒ x^{2} (1 – m^{2}) + 2mxy = 0.
Which represents the pair of lines between which the angle is 45^{o}.
⇒ 1 – m^{2 }= ± 2m ⇒ m^{2 }± 2m – 1 = 0
The centres of a set of circles, each of radius 3, lie on the circle x^{2}+y^{2}=25. The locus of any point in the set is
For any point P (x, y) in the given circle,
we should have
OA ≤ OP ≤ OB ⇒ (53)
⇒ 4 ≤ x^{2} +y^{2} ≤ 64
The centre of the circle passing through (0, 0) and (1, 0) and touching the circle x^{2}+y^{2}=9 is [2002]
Let the required circle be x^{2} + y^{2} + 2gx + 2fy + c = 0
Since it passes through (0, 0) and (1, 0)
⇒ c = 0 and g = 
Points (0, 0) and (1, 0) lie inside the circle x^{2} + y^{2} = 9, so two circles touch internally
⇒ c_{1}c_{2} = r_{1} – r_{2}
Hence, the centres of required circle are
The equation of a circle with origin as a centre and passing through equilateral triangle whose median is of length 3a is
Let ABC be an equilateral triangle, whose median is AD.
Given AD = 3a.
In Δ ABD, AB^{2} = AD^{2} + BD^{2} ; ⇒ x^{2} = 9a^{2} + (x^{2}/4)
where AB = BC = AC = x.
In Δ OBD, OB^{2} = OD^{2} + BD^{2}
⇒ r^{2} = 9a^{2 }– 6ar + r^{2} + 3a^{2} ;
⇒ 6ar = 12a^{2} ⇒ r = 2a
So equation of circle is x^{2} + y^{2} = 4a^{2}
If the two circles ( x  1)^{2} + ( y  3)^{2} = r^{2 }and x^{2} + y^{2}  8 x + 2 y + 8 = 0 intersect in two distinct point, then[2003]
r_{1}  r_{2}< C_{1}C_{2} for intersection
and ,
From (1) and (2), 2 < r < 8.
The lines 2 x  3y=5 and 3x  4 y = 7 are diameters of a circle having area as 154 sq.units.Then the equation of the circle is[2003]
πr^{2} = 154 ⇒r= 7
For centre on solving equation 2x 3y = 5& 3x 4y= 7
we get x = 1,y =1
∴ centre = (1, –1 )
Equation of circle, ( x  1)^{2} + (y + 1)^{2}
=72 x^{2} + y^{2} 2x+2y= 47
If a circle passes through the point (a, b) and cuts the circle x^{2} + y^{2} = 4 orthogonally, then the locus of its centre is
Let the variable circle is x^{2 }+ y^{2} + 2gx + 2fy + c = 0 ......(1)
It passes through (a, b)
∴ a^{2} + b^{2} + 2 ga + 2 fb + c = 0 ......(2)
(1) cuts x^{2} +y^{2}= 4 orthogonally
∴2(gx0+ f x0) = c4 ⇒ c=4
∴ from (2) a^{2} + b^{2} + 2ga + 2fb + 4=0
∴ Locus of centre (–g,–f) is a^{2} + b^{2}  2ax 2by +4= 0
or 2ax + 2by = a^{2} + b^{2 }+ 4
A variable circle passes through the fixed point A( p,q) and touches xaxis . The locus of the other end of the diameter through A is [2004]
Let the variable circle be x^{2} + y^{2} + 2gx + 2fy + c = 0 ....(1)
∴ p^{2} + q^{2} + 2gp + 2fq + c =0 ....(2)
Circle (1) touches xaxis,
∴ g^{2}  c = 0 ⇒ c=g^{2} .
From (2) p^{2 }+ q^{2} + 2gp + 2fq +g^{2 }=0 ....(3)
Let the other end of diameter through (p, q) be (h, k),
then
Put in (3)
⇒ h^{2 }+ p^{2}  2hp  4kq=0
∴ locus of (h, k) is x^{2} + p^{2}  2 xp  4 yq=0
⇒ ( x  p )2=4qy
If the lines 2x + 3 y + 1 = 0 and 3x y  4=0 lie alon g diameter of a circle of circumference 10p, then the equation of the circle is [2004]
Two diameters are along 2x + 3y +1 = 0 and 3x y  4=0
solving we get centre (1, –1) circumference = 2πr = 10π
∴ r= 5.
Required circle is, ( x  1)^{2} + (y + 1)^{2}=5^{2}
⇒ x^{2} + y^{2} 2x+ 2y 23=0
Intercept on the line y = x by the circle x^{2} + y^{2}  2x = 0 is AB. Equation of the circle on AB as a diameter is [2004]
Solving y = x and the circle x^{2} + y^{2 } 2x= 0,
we get x = 0, y = 0 and x = 1,y=1
∴ Extremities of diameter of the required circle are (0, 0) and (1, 1).
Hence, the equation of circle is (x  0)(x 1) + (y  0)(y 1)= 0
⇒ x^{2} + y^{2}  x y= 0
If th e circles x^{2} + y^{2 }+ 2ax + cy + a = 0 and x^{2} + y^{2} – 3ax + dy – 1 = 0 intersect in two distinct points P and Q then the line 5x + by – a = 0 passes through P and Q for[2005]
s_{1} = x^{2} + y^{2}+ 2ax + cy +a =0
s_{2} = x^{2 }+ y^{2 } 3ax + dy  1 =0
Equation of common chord of circles s_{1} and s_{2} is given by s_{1 } s_{2}= 0
⇒ 5ax + (c  d)y +a +1=0
Given that 5x + by – a = 0
passes through P and Q
∴ The two equations should represent the same line
a^{2 }+a+1=0
No real value of a.
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[2005]
Equation of circle with centre (0, 3) and radius 2 is x^{2} + (y  3)^{2} = 4
Let locus of the variable circle is (α , β)
∵ It touches x  axis.
∴ It's equation is ( x  α)^{2} + (y + β)^{2 }= β^{2}
Circle touch externally ⇒ c_{1}c_{2} = r_{1 }+ r_{2}
∴ Locus is which is a parabola.
If a circle passes through the point (a, b) and cuts the circle x^{2} + y^{2 }=^{ }p^{2 }orthogonally, then the equation of the locus of its centre is [2005]
Let the centre be (α, β)
∵ It cuts the circle x^{2} + y^{2 }= p^{2} orthogonally
∴ Using 2 g_{1} g_{2} +2 f_{1} f_{2} = c_{1}+c_{2} , we get
2(α) x 0 + 2(β) x 0 = c_{1} p^{2} ⇒ c_{1} = p^{2}
Let equation of circle is x^{2} + y^{2}  2αx 2βy+p^{2 }=0
It passes through (a, b) ⇒ a^{2} + b^{2}  2αa  2βb +p^{2}=0
∴Locus of (α, β) is
∴2ax+ 2by  (a^{2} +b^{2 }+ p^{2})=0 .
If the pair of lines ax^{2} + 2 (a + b)xy + by 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 [2005]
As per question area of one sector = 3 area of another sector
⇒ angle at centre by one sector = 3x angle at centre by another sector
Let one angle be θ then other = 3θ
Clearly θ + 3θ = 180 ⇒ θ = 45^{o}
∴ Angle between the diameters represented by combined equation
ax^{2} + 2 ( a + b ) xy + by^{2 }= 0 is 45^{o}
∴ Using
we get
⇒ a^{2} + b^{2} + 2ab = 4a^{2} + 4b^{2 }+4ab ⇒ 3a^{2 }+ 3b^{2} + 2ab = 0
If the lines 3x  4y  7=0 and 2x  3y  5=0 are two diameters of a circle of area 49π square units, the equation of the circle is [2006]
Point of intersection of 3x  4y  7=0 and 2 x  3y  5=0 is (1,  1)
which is the centre of the circle and radius = 7
∴ Equation is ( x  1)^{2} + (y + 1)^{2}= 49
⇒ x^{2 }+ y^{2 }2x+ 2y47=0
Let C be the circle with centre (0, 0) and radius 3 units. The equation of the locus of the mid points of the chords of the circle C that subtend an angle of at its center is
Let M(h, k) be the mid point of chord AB where
∴ Locus of (h, k) is
Consider a family of circles which are passing through the point (– 1, 1) and are tangent to xaxis. If (h, k) are the coordinate of the centre of the circles, then the set of values of k is given by the interval [2007]
Equation of circle whose centre is (h, k)
i.e (x – h)^{2} + (y – k)^{2} = k^{2}
(radius of circle = k because circle is tangent to xaxis) Equation of circle passing through (–1, +1)
∴ (–1 –h)^{2} + (1 – k)^{2} = k^{2}
⇒ 1 + h^{2} + 2h + 1 + k^{2} – 2k = k^{2}
⇒ h^{2} + 2h – 2k + 2 = 0 D ≥ 0
∴ (2)^{2} – 4 × 1.(–2k + 2) ≥ 0 ⇒ 4 – 4(–2k + 2) ≥ 0 ⇒ 1 + 2k – 2 ≥ 0 ⇒
The point diametrically opposite to the point P(1, 0) on the circle x^{2} + y^{2} + 2x + 4y – 3 = 0 is [2008]
The given circle is x^{2} + y^{2} + 2x + 4y –3 = 0
Centre (–1, –2)
Let Q (α, β) be the point diametrically opposite to the point P(1, 0),
⇒ α = –3, β = – 4, So, Q is (–3, –4)
The differential equation of the family of circles with fixed radius 5 units and centre on the line y = 2 is
Let the centre of the circle be (h, 2)
∴ Equation of circle is ( x– h)^{2} + (y – 2)^{2} = 25 …(1)
Differentiating with respect to x, we get
Substituting in equation (1) we get
⇒ (y – 2)^{2} (y')^{2} = 25 – (y –2)^{2}
If P an d Q are the points of intersection of the circles x^{2} + y^{2 }+ 3x + 7y+ 2p 5 =0 and x^{2 }+ y^{2} + 2x + 2y – p^{2 }= 0 then there is a circle passing through P, Q and (1, 1) for: [2009]
The given circles are S_{1} ≡ x^{2} + y^{2} + 3x + 7y + 2p – 5 = 0 ....(1)
S_{2} ≡ x^{2 }+ y^{2} + 2x + 2y – p^{2} = 0 ....(2)
∴ Equation of common chord PQ is S_{1} – S_{2} = 0
⇒ L≡ x+5y+ p^{2} + 2p 5=0
⇒ Equation of circle passing through P and Q is S_{1} + λL = 0
⇒ (x^{2} + y^{2} + 3x + 7y + 2p – 5) + λ (x + 5y + p^{2} +2p – 5) = 0
As it passes through (1, 1), therefore
⇒ (7 + 2p ) + λ (2p + p^{2 }+ 1) = 0
which does not exist for p = – 1
The circle x^{2} + y^{2} = 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if [2010]
Circle x^{2 }+ y^{2}  4 x  8y  5=0
Centre = (2, 4), Radius =
If circle is intersecting line 3x  4 y=m, at two distinct points.
⇒ length of perpendicular from centre to the line < radius
⇒ 10 + m < 25
⇒ –25 < m + 10 < 25 ⇒ – 35 < m < 15
The two circles x^{2 }+ y^{2} = ax and x^{2 }+ y^{2} = c^{2} (c > 0) touch each other if [2011]
As centre of one circle is (0, 0) and other circle passes through (0, 0), therefore
Also
If the two circles touch each other, then they must touch each other internally.
The length of the diameter of the circle which touches the xaxis at the point (1,0) and passes through the point (2,3) is: [2012]
Let centre of the circle be (1,h)
Let the circle passes through the point B (2,3)
∴ CA = CB (radius) ⇒ CA^{2} = CB^{2}
⇒ (1 – 1)^{2} + (h – 0)^{2 }= (1 – 2)^{2} + ( h – 3)^{2 }
⇒ h^{2} = 1 + h^{2} + 9 – 6h ⇒
Thus, diameter is
The circle passing through (1, –2) and touching the axis of x at (3, 0) also passes through the point [JEE M 2013]
Since circle touches xaxis at (3, 0)
∴ The equation of circle be (x – 3)^{2} + (y – 0)^{2} + λy = 0
As it passes through (1, –2)
∴ Put x = 1, y = –2
⇒ (1 – 3)^{2 }+ (–2)^{2} + λ(–2) = 0 ⇒ λ = 4
∴ equation of circle is (x – 3)^{2} + y^{2} – 8 = 0
Now, from the options (5, –2) satisfies equation of circle.
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 [JEE M 2014]
Equation of circle C ≡ ( x 1)^{2} + (y 1)^{2} =1
Radius of T =  y 
T touches C externally therefore,
Distance between the centres = sum of their radii
⇒ (0 – 1)^{2 }+ (y –1)^{2} = (1 + y)^{2 }
⇒ 1 + y^{2} + 1 – 2y = 1 + y^{2} + 2 y 
2  y  = 1 – 2y
If y > 0 then 2y = 1 – 2y ⇒ y =
If y < 0 then –2y = 1 – 2y ⇒ 0 = 1 (not possible)
∴ y =
Locus of the image of the point (2, 3) in the line (2x – 3y + 4) + k (x – 2y + 3) = 0, k ∈ R, is a : [JEE M 2015]
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
(α – 1)^{2} + (β – 2)^{2} = (2 – 1)^{2} + (3 – 2)^{2 }
(α – 1)^{2} + (β –2)^{2} = 1 + 1 = 2
(x – 1)^{2} + (y – 2)^{2} =
(x – α)^{2} + (y – β)^{2} = r^{2 }
Therefore, given locus is a circle with centre (1, 2) and radius
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 :[JEE M 2015]
x^{2} + y^{2} – 4x – 6y – 12 = 0 ...(i)
Centre, c_{1} = (2, 3) and
Radius, r_{1} = 5 units x^{2} + y^{2} + 6x + 18y + 26 = 0 ...(ii)
Centre, c_{2} = (–3, –9) and Radius, r_{2} = 8 units
r_{1 }+ r_{2 }= 5 + 8 = 13
∴ C_{1} C_{2} = r_{1} + r_{2}
Therefore there are three common tangents.
The centres of those circles which touch the circle, x^{2} + y^{2} – 8x – 8y – 4 = 0, externally and also touch the xaxis,lie on: [JEE M 2016]
For the given circle, centre : (4, 4) radius = 6
(h – 4)2 = 20k + 20
∴ locus of (h, k) is (x – 4)^{2} = 20(y + 1), which is a parabola.
If one of the diameters of the circle, given by the equation, x^{2} + y^{2 }– 4x + 6y – 12 = 0, is a chord of a circle S, whose centre is at (–3, 2), then the radius of S is: [JEE M 2016]
Centre of S : O (–3, 2) centre of given circle A(2, –3)
Also AB = 5 (∵ AB = r of the given circle)
⇒ Using pythagoras theorem in ΔOAB
130 videos359 docs306 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
130 videos359 docs306 tests









