1 Crore+ students have signed up on EduRev. Have you? 
A square is inscribed in the circle x^{2 }+ y^{2 }– 2x + 4y + 3 = 0. Its sides are parallel to the coordinate axes. The one vertex of the square is (1980)
The given circle is x^{2} + y^{2} – 2x + 4y + 3 = 0. Centre (1, – 2). Lines through centre (1, – 2) and parallel to axes are x = 1 and y = – 2.
Let the side of square be 2k.
Then sides of square are x = 1 – k and x = 1 + k and y = – 2 – k and y = – 2 + k
∴ Coordinates of P, Q, R, S are (1 + k, – 2 + k), (1 – k, – 2 + k), (1 – k, – 2 – k), (1 + k, – 2 – k) respectively.
Also P (1 + k, – 2 + k) lies on circle
∴ (1 + k)^{2} + (– 2 + k)^{2} – 2 (1 + k) + 4 (–2 + k) + 3 = 0 ⇒ 2k^{2} = 2
⇒ k = 1 or – 1 If k = 1, P (2, – 1), Q (0, – 1), R (0, – 3), S (2, – 3)
If k = – 1, P (0, – 3), Q (2, – 3), R (2, – 1), S(0, – 1)
Two circles x^{2} + y^{2} = 6 and x^{2} + y^{2 }– 6x + 8 = 0 are given. Then the equation of the circle through their points of intersection and the point (1, 1) is (1980)
The circle through points of intersection of the two circles x^{2} + y^{2 }– 6 = 0
and x^{2} + y^{2} – 6x + 8 = 0
is (x^{2} + y^{2} – 6) + l (x^{2} + y^{2} – 6x + 8) = 0
As it passes through (1, 1) (1 + 1 – 6) + l (1 + 1 – 6 + 8) = 0 ⇒
∴ The required circle is 2x^{2} + 2y^{2} – 6x + 2 = 0
or, x^{2} + y^{2} – 3x + 1 = 0
The centre of the circle passing through the point (0, 1) and touching the curve y = x^{2} at (2, 4) is (1983  1 Mark)
Let C (h, k) be the centre of circle touching x^{2} = y at B (2, 4). Then equation of common tangent at B is
i.e., 4x – y = 4
Radius is perpendicular to this tangent
= 1 ⇒ 4k = 18… (1)
Also AC = BC ⇒ h^{2} + (k – 1)^{2 }= (h – 2)^{2} + (k – 4)^{2 }
⇒ 4h + 6k = 19 … (2)
Solving (1) and (2) we get the centre as
The equation of the circle passing through (1, 1) and the points of intersection of x^{2} + y^{2} + 13x – 3y = 0 and 2x^{2} + 2y^{2} + 4x – 7y – 25 = 0 is (1983  1 Mark)
KEY CONCEPT
Circle through pts. of intersection of two circles S_{1 }= 0 and S_{2} = 0 is S_{1} + λS_{2} = 0
∴ Req. circle is, (x^{2} + y^{2} + 13x – 3y) + λ (x^{2} + y^{2 }+ 2x
⇒ (1 + λ) x^{2} + (1 +λ) y^{2} + (13 + 2λ)
As it passes through (1, 1)
∴ 1 + λ + 1 + λ + 13 + 2λ – 3 –
⇒ – 12λ + 12 = 0 ⇒ λ = 1
∴ Req. circle is,
2x^{2} + 2y^{2} + 15 x 
or4x^{2} + 4y^{2 }+ 30x –13 y – 25 = 0
The locus of the mid point of a chord of the circle x^{2} + y^{2} = 4 which subtends a right angle at the origin is (1984  2 Marks)
Let AB be the chord with its mid pt M (h, k ).
As ∠AOB = 90°
∴ AM =
NOTE THIS STEP
By prop. of rt. Δ AM = MB = OM
∴ OM = ⇒ h^{2} +k^{2 }= 2
∴ locus of (h, k) is x^{2} + y^{2} = 2
If a circle passes through the point (a, b) and cuts the circle x^{2} + y^{2} = k^{2 }orthogonally, then the equation of the locus of its centre is (1988  2 Marks)
KEY CONCEPT
Two circles x^{2} + y^{2 }+ 2g_{1}x + 2f_{1}y + c_{1} = 0 and
x^{2 }+ y^{2} + 2g_{2} x + 2f_{2} y + c_{2} = 0 are orthogonal iff
2g_{1}g_{2} + 2f_{1} f_{2} = c_{1} + c_{2}
(a) Let the required circle be, x^{2} + y^{2} + 2gx + 2fy + c = 0 … (1)
As it passes through (a, b), we get,
a^{2} + b^{2} + 2ag + 2bf + c = 0 … (2)
Also (1) is orthogonal with the circle, x^{2} + y^{2} = k^{2} … (3)
For circle (1) g_{1} = g, f_{1} = f, c_{1} = c
For circle (3) g_{2} = 0, f_{2 }= 0, c_{2} = – k^{2 }
∴ By the condition of orthogonality,
2g_{1}g_{2} + 2f_{1}f_{2} = c_{1} + c_{2} We get, c = k^{2}
Substituting this value of c in eq. (2), we get
a^{2} + b^{2} + 2ga + 2f b + k^{2} = 0
∴ Locus of centre (g – f) of the circle can be obtained by replacing g by –x and f by – y
we get a^{2} + b^{2 }– 2ax – 2by + k^{2} = 0
i.e.2ax + 2by – (a^{2 }+ b^{2} + k^{2}) = 0
If the two circles (x – 1)^{2} + (y – 3)^{2} = r^{2} and x^{2} + y^{2} – 8x + 2y + 8 = 0 intersect in two distinct points, then (1989  2 Marks)
We have two circles (x – 1)^{2} + (y – 3)^{2} = r^{2}
Centre (1, 3), radius = r and x^{2} + y^{2} – 8x + 2y + 8 = 0
Centre (4, – 1), radius =
As the two circles intersect each other in two distinct points we should have
C_{1 }C_{2} < r_{1 }+ r_{2} and C_{1} C_{2} >  r_{1} – r_{2} 
⇒ C_{1 }C_{2} < r + 3 and C_{1} C_{2} <  r_{1} – r_{2}
and 5 > r – 3 
⇒ 5 < r + 3 ⇒  r – 3  < 5 ⇒ r > 2 … (i)
⇒ – 5 < r – 3 < 5 ⇒ – 2 < r < 8 … (ii)
Combining (i) and (ii), we get 2 < r < 8
The lines 2x – 3y = 5 and 3x – 4y = 7 are diameters of a circle of area 154 sq. units. Then the equation of this circle is
As 2x – 3y – 5 = 0 and 3x – 4y – 7 = 0 are diameters of circles.
∴ Centre of circle is solution of these two eq. ' s, i.e.
⇒ x = 1, y = – 1
∴ C (1, –1)
Also area of circle, πr^{2} = 154
∴ Equation of required circle is
(x – 1)^{2} + (y + 1)^{2} = 72
⇒ x^{2} + y^{2} – 2x + 2y = 47
The centre of a circle passing through the points (0, 0), (1, 0) and touching the circle x^{2} + y^{2} = 9 is (1992  2 Marks)
Let the equation of the circle be
x^{2} + y^{2} + 2gx + 2fy + c = 0.
As this circle passes through (0, 0) and (1, 0)
we get c = 0, 1 + 2g = 0
⇒ g =
According to the question this circle touches the given circle x^{2} + y^{2} = 9
∴ 2 × radius of required circle = radius of given circle
∴ The centre is
The locus of the centre of a circle, which touches externally the circle x^{2} + y^{2 }– 6x – 6y + 14 = 0 and also touches the yaxis, is given by the equation: (1993  1 Marks)
The given circle is x^{2} + y^{2} – 6x + 14 = 0, centre (3, 3), radius = 2
Let (h, k) be the centre of touching circle.
Then radius of touching circle = h [as it touches yaxis also]
∴ Distance between centres of two circles = sum of the radii of two circles
⇒ (h – 3)^{2} + (k – 3)^{2} = (2 + h)^{2}
⇒ k^{2} – 10h – 6k + 14 = 0
∴ locus of (h, k) is y^{2 }– 10x – 6y + 14 = 0
The circles x^{2} + y^{2} – 10x + 16 = 0 and x^{2} + y^{2} = r^{2} intersect each other in two distinct points if (1994)
Centres and radii of two circles are C_{1} (5, 0); 3 = r_{1} and C_{2} (0, 0); r = r_{2}
As circles intersect each other in two distinct points,
r_{1} – r_{2} < C_{1}C_{2} < r_{1} + r_{2 }
⇒  r – 3  < 5 < r + 3 ⇒ 2 < r < 8
The angle between a pair of tangents drawn from a point P to the circle x^{2} + y^{2} + 4x – 6y + 9 sin^{2}α + 13 cos^{2}α = 0 is 2α.The equation of the locus of the point P is (1996  1 Mark)
Centre of the circle x^{2} + y^{2} + 4x – 6y + 9 sin^{2}α + 13 cos^{2}α = 0 is C (– 2, 3) and its radius is
Let P (h, k) be any point on the locus. The ∠ APC =α
Also ∠ PAC =
That is, triangle APC is a right triangle.
Thus,
⇒ (h + 2)^{2} + (k – 3)^{2} = 4 or h^{2 }+ k^{2} + 4h – 6k + 9 = 0
Thus required equation of the locus is x^{2 }+ y^{2} + 4x – 6y + 9 = 0
If two distinct chords, drawn from the point (p, q) on the circle x^{2} + y^{2 }= px + qy (where pq ≠ 0) are bisected by the x –axis, then (1999  2 Marks)
The given equation of the circle is x^{2} + y^{2} – px – qy = 0, pq ≠ 0 Let the chord drawn from (p, q) is bisected by xaxis at point (x1, 0).
Then equation of chord is
(using T = S_{1})
As it passes through (p, q), therefore,
As through (p,q) two distinct chords can be drawn.
∴ Roots of above equation be real and distinct, i.e., D > 0.
⇒ 9p^{2 }– 4 × 2 (p^{2} + q^{2}) > 0 ⇒ p^{2} > 8q^{2}
The triangle PQR is inscribed in the circle x^{2} + y^{2} = 25. If Q and R have coordinates (3,4) and (–4, 3) respectively, then ∠QPR is equal to (2000S)
O is the point at centre and P is the point at circumference. Therefore, angle QOR is double the angle QPR.
So, it sufficient to find the angle QOR. Now slope of OQ = 4/3
Slope of OR = – 3/4
Again m_{1} m_{2} = – 1
Therefore, ∠QOR = 90° which implies that ∠QPR = 45°.
If the circles x^{2} + y^{2} + 2x + 2ky + 6 = 0, x^{2} + y^{2} + 2ky + k = 0 intersect orthogonally, then k is (2000S)
2g_{1}g_{2} + 2f_{1}f_{2} = c_{1}+ c_{2} (formula for orthogonal intersection of two cricles)
⇒ 2 (1) (0) + 2 (k) (k) = 6 + k
⇒ 2k^{2 }– k – 6 = 0
⇒ k = – 3/2, 2
Let AB be a chord of the circle x^{2} + y^{2} = r^{2} subtending a right angle at the centre. Then the locus of the centroid of the triangle PAB as P moves on the circle is (2001S)
x^{2 }+ y^{2} = r^{2} is a circle with centre at (0, 0) and radius r units.
Any arbitrary pt P on it is (r cosθ , r sinθ ) Choosing A and B as (– r, 0) and (0, – r).
[So that ∠AOB = 90°]
For locus of centriod of ΔABP
⇒ r cosθ – r = 3x
r sinθ – r = 3y
⇒ r cosθ = 3x + r
r sinθ = 3y + r
Squaring and adding (3x + r)^{2} + (3y + r)^{2} = r^{2} which is a circle.
Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at a point X on the circumference of the circle, then 2r equals (2001S)
Let ∠RPS = θ
∠XPQ = 90° – θ
∴∠PQX = θ (∵ ∠PXQ = 90°)
∴ΔPRS ~ ΔQPR (AA similarity)
If the tangent at the point P on the circle x^{2} + y^{2} + 6x + 6y = 2 meets a straight line 5x –2y + 6 = 0 at a point Q on the y  axis, then the length of PQ is (2002S)
Line 5x – 2y + 6 = 0 is intersected by tangent at P to circle x^{2} + y^{2} + 6x + 6y – 2 = 0 on yaxis at Q (0, 3).
In other words tangent passes through (0, 3)
∴ PQ = length of tangent to circle from (0, 3)
The centre of circle inscribed in square formed by the lines x^{2} – 8x + 12 = 0 and y^{2 }– 14y + 45 = 0, is (2003S)
x^{2} – 8x + 12 = 0 ⇒ (x – 6) (x – 2) = 0
y^{2} – 14y + 45 = 0 ⇒ (y – 5) (y – 9) = 0
Thus sides of square are x = 2, x = 6, y = 5, y = 9
Then centre of circle inscribed in square will be
If one of the diameters of the circle x^{2} + y^{2} – 2x – 6y + 6 = 0 is a chord to the circle with centre (2, 1), then the radius of the circle is (2004S)
The given circle is x^{2} + y^{2} – 2x – 6y + 6 = 0 with centre C(1, 3) and radius
Let AB be one of its diameter which is the chord of other circle with centre at C_{1} (2, 1).
Then in ΔC_{1}CB,
r^{2} = [(2 – 1)^{2} + (1 – 3)^{2}] + (2)^{2}
⇒ r^{2} = 1 + 4 + 4 ⇒ r^{2 }= 9 ⇒ r = 3.
A circle is given by x^{2} + (y–1)^{2} = 1, another circle C touches it externally and also the xaxis, then the locus of its centre is (2005S)
Let the centre of circle C be (h, k). Then as this circle touches axis of x, its radius =  k 
Also it touches the given circle x^{2 }+ (y – 1)^{2} = 1, centre (0, 1) radius 1, externally
Therefore, the distance between centres = sum of radii
⇒ h^{2} + k^{2} – 2k + 1 = (1 +  k )^{2 }
⇒ h^{2} + k^{2} – 2k + 1 = 1 + 2  k  + k^{2 }
⇒ h^{2} = 2k + 2  k 
∴ Locus of (h, k) is, x^{2} = 2y + 2  y 
Now if y > 0, it becomes x^{2} = 4y and if y < 0, it becomes x = 0
∴ Combining the two, the required locus is {(x, y) : x^{2 }= 4y} ∪ {(0, y) : y < 0}
Tangents drawn from the point P(1, 8) to the circle x^{2 }+ y^{2} –6x– 4y –11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is (2009)
Tangents PA and PB are drawn from the point P (1, 3) to circle x^{2} + y^{2 } 6 x  4y  11=0 with centre C (3, 2)
Clearly the circumcircle of ΔPAB will pass through C and as ∠A = 90°, PC must be a diameter of the circle.
∴ Equation of required circle is (x 1)(x  3) + (y8)(y  2)= 0
⇒ x^{2} + y^{2 } 4x  10y + 19= 0
The circle passing through the point (– 1, 0) and touching the yaxis at (0, 2) also passes through the point. (2011)
Let centre of the circle be (h, 2) then radius = h
∴ Equation of circle becomes (x – h)^{2 }+ (y – 2)^{2} = h^{2}
As it passes through (–1, 0)
⇒ (–1 – h)^{2} + 4 = h^{2 }⇒ h =
Distance of centre from (–4, 0) is
∴ It lies on the circle.
The locus of th e midpoint of the ch or d of con tact of tangents drawn from points lying on the straight line 4x – 5y = 20 to the circle x^{2} + y^{2 }= 9 is (2012)
Any point P on line 4x – 5y = 20 is
Equation of chord of contact AB to the circle x^{2} + y^{2 }= 9
drawn from point
....(1)
Also the equation of chord AB whose mid point is (h, k) is
hx + ky = h^{2}+k^{2} ....(2)
∵ Equations (1) and (2) represent the same line, therefore
⇒ 5kα= 4hα 20h and 9h = α( h^{2}+k^{2})
and
∴ Locus of (h, k) is
(20 x^{2 }+ y^{2})  36x + 45y= 0
132 docs70 tests

Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
132 docs70 tests








