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Transcript of Circles Imp
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LAQ (2 )1) Find the equation of the equation of the circle
passing through the points
1. (1, 1), (2, -1), (3, 2)2. (5, 7), (8, 1), (1, 3)3. (1, 2), (3, -4), (5, -6)
Sol: let the equation of the required circle be
x2+y2+2gx+2fy+c=0. (1, 1) lies on S=0
(1)2+ (1)2+2g (1) +2f (1) +c=02g + 2f +c =-2..
(2, -1) lies on S=0
(2)2+ (-1)2+2g (2) +2f (-1) +c=04g - 2f +c =-5..
(3, 2) lies on S=0
(3)2+ (2)2+2g (3) +2f (2) +c=06g + 4f +c =-13.
Solving eqn & & 2g + 2f +c =-2 4g - 2f +c =-5
4g - 2f +c =-5 6g + 4f +c =-13
-2g+4f =3 -2g-6f=8.
Solving eqn & sub f value in eqn
-2g+4f =3 -2g+4() =3
-2g-6f=8 -2g=3+2
10f=-5 g=
f=-Sub the value of g and f in eqn
2 ( ) + 2(-) +c =-2
-5-1+c=-2 c=4
the required eqn of the circle isx2+y2-5x-y+4=0.
{Ans: of 2 and 3 3(x2+y2)-29x-19y+56=0,x2+y2-22x-4y+25=0,}
2) Show that the four points (1, 1), (-6, 0), (-2, 2), and(-2, -8) are concyclic and find the equation of the
circle on which they lie.
{H/W (ii)(9,1), (7, 9), (-2, 12), &(6, 10)}
Sol: let the equation of the required circle be
x2+y2+2gx+2fy+c=0. A (1, 1) lies on S=0
(1)2+ (1)2+2g (1) +2f (1) +c=0
2g + 2f +c =-2..
B (-6, 0) lies on S=0
(-6)2+ (0)2+2g (-6) +2f (0) +c=0-12g +c =-36..
C (-2, 2) lies on S=0
(-2)2+ (2)2+2g (-2) +2f (2) +c=0-4g + 4f +c = - 8.
Solving eqn & & 2g + 2f +c =-2 -12g +0 +c =-36
-12g +c =-36 -4g + 4f +c =-8
14g+2f =34 -8g-4f=-28.
Solving eqn & sub f value in eqn 14g+2f =34 14g+2f =34
4g+2f=14 14(2) +2f =3410g=20 2f=-28
g=2 f=3
Sub the value of g and f in eqn 2 () + 2( ) +c =-24+6+c=-2c=-12
the required eqn of the circle isx2+y2+4x+6y-12=0.
Now substituting D (-2, -8) in the above eqn, we have(-2)2+(-8)2+4(-2)+6(-8)-12
=4+64-8-48-12
=68-68
=0
D (-2, -8) lies on the circle given 4 points are concyclic.
3) If (2, 0), (0, 1), (4, 5) and (0, c) are concyclic, andthen find the value of c.{H/w (1, 2), (3, -4), (5, -6), (c,
8)}
Sol: let the equation of the required circle be
x2+y2+2gx+2fy+c=0. A (2, 0) lies on S=0
(2)2+ (0)2+2g (2) +2f (0) +c=04g +k=-4..
B (0, 1) lies on S=0
(0)2+ (1)2+2g (0) +2f (1) +c=0 2f +k =-1..
C (4, 5) lies on S=0
(4)2+ (5)2+2g (4) +2f (5) +c=08g + 10f +k =-41.
Solving eqn & &
4g + 0 + k =-4 0 + 2f + k =-10 + 2f +k =-1 8g + 10f +k =-41
4g - 2f =-3 -8g-8f=40.
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Solving eqn & sub f value in eqn
4g-2f =-3 4g-2( ) =-3-4g-4f=20 4g=-3-
-6f=17 g=
f=- g=- Sub the value of g and f in eqn 4g +k=-4
4 (- ) +k=-4
the required eqn of the circle is
x2+y2+2( )x+2( )y+ =0.3x2+3y2-13x-17y+14=0 given 4 points are concyclic,D (0, c) lies on the above circle
3(0)2+3(c)2-13(0)-17(c)+14=03c2-17c+14=03c2-3c-14c+14=03c(c-1)-14(c-1) =0(c-1) (3c-14) =0(c-1) =0 or (3c-14) =0
c=1, c=
4) Find the equation of the equation of the circlepassing through the points (4, 1), (6, 5) and whosecentre lies on {H/W (2, -3), (-4, 5) and 4x+3y+1=0, (4, 1), (6, 5),
and 4x+y-16=0}
Sol: let the equation of the required circle be
x2+y2+2gx+2fy+c=0. (4, 1) lies on S=0
(4)2+ (1)2+2g (4) +2f (1) +c=08g + 2f +c =-17..
(6, 5) lies on S=0 (6)2+ (5)2+2g (6) +2f (5) +c=0
12g +10f+c =-61..Solving eqn & 8g + 2f +c =-17
12g +10f+c =-61
-4g-8f = 44.Given centre (-g, -f) lies on 4x+3y-16=0
4 (-g) +3(-f)-24=04g+3f+24=0
Solving eqn &
-4g-8f = 44
4g+3f=-24
-5f = 20 f=-4
From eqn 4g+3f+24=04g+3(-4)=-24
4g=-24+124g=-12g=-3From (1)
8g + 2f +c =-178(-3)+2(-4)+c=-17c=-17+24+8C=15
Required eqn of the circle is x2+y2-6x-8y+15=0.
5) Find the equation of the circles whose centre lieson X-axis and passing through (-2, 3) and (4, 5).
Sol: Sol: let the equation of the required circle be
x2+y2+2gx+2fy+c=0. (-2, 3) lies on S=0
(-2)2
+ (3)2
+2g (-2) +2f (3) +c=0-4g + 6f +c =-13..
(4, 5) lies on S=0
(4)2+ (5)2+2g (4) +2f (5) +c=08g +10f+c =-41..Solving eqn & -4g + 6f +c =-13
8g +10f+c =-41
-12g-4f = 28.Given centre (-g, -f) lies on X- axis
-f=0 or f=0-12g-0=28
g=-
Substituting f=0,g=-7/3 in eqn (1)-4g + 6f +c =-13
-4() +6(0) +c=-13
C=-13- C= -
Required eqn of the circle is
x2+y2+2(
)x+2(0)y+(
)=0.
3 x2+3y2-14x-67=0.
6) Show that the circles touch each other. Also findthe point of contact and common tangent at this
point of contact.
a) x2+y2-6x-2y+1=0,x2+y2+2x-8y+13=0.
Sol: given equation of the circles
Sx2+y2-6x-2y+1=0, and
Centre C1= (3, 1) C2 (-1, 4)
And r=
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r1=() () () ()
() ( ) =5 given two circles touch externally
The point of contact P divides i th rti r r P=0 1=0()()
()() 1
=0
1 0
1Equation of common tangent at point of contact is the
radical axis of the circles S-S=0
(x2+y2-6x-2y+1=0)- (x2 ) =0-8x+6y-12=0
Or 4x-3y+6=0.
7) Find the equation of the pair of transversecommon tangents to the circles
x2+y2-4x-10y+28=0,x2+y2+4x-6y+4=0.
Sol: Sol: given equation of the circles
Sx2+y2-4x-10y+28=0, and
Centre C1= (2, 5) C2 (-2, 3)
And r=
r1=() () () ()
() ( ) = There exist two transverse common tangents
Internal centre of similitude P divides i th rti r r
P=0 1=0()()
()()
1
=0 1 0 1 0 1The equation to the pair of transverse common tangents
through P(1,9/2) to the circle S
is ( ) ( ) Where ( ) =(1,
)
0() ./ ( ) . / 1
=, - 0() () () ./ 1
0 1
=, - 0 1
0 1=, - 01
01 , -=, - 01, -=, - {4 }
=, -3 =0
x2+y2-4x-6y-12=0, x2+y2+6x+18y+26=0.(H/W)
8) Find the equation of direct common tangents to thecircles x2+y2+22x-4y-100=0,
x2+y2-22x+4y+100=0.
Sol: given equation of the circles
Sx2+y2+22x-4y-100=0, and Centre C1= (-11, 2) C2 (11,-2)
And r= r1=() ()
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() ()
() ( ) =
There exist two direct common tangents
External centre of similitude Q divides i th rtir r
Q=0 1=0()()
()()
1
=0 1 0 1,-The equation to the pair of direct common tangents
through Q(22, -4) to the circle S is ( ) ( ) Where( )=(22, -4) ,() () ( ) ( ) -=, -,() () ()()-, -=, -, -,-=, -,-, -=, -,-, -=, -,-{121 }
=, -21 =0
9) Find the equation of the circles which touch2x-3y+1=0 at (1, 1) and having radius.
Sol: given two circles touch the line
( ) ( )
() The centres
() ( ) ()Slope of (1) = - (
) =2
Slope of line
The centres =. 0 1 0
1/
= {1()}= (1-2, 1+3) and (1+2, 1-3)
= (-1, 4) and (3, -2)
Eqn of the circle with
()
( )()
Eqn of the circle with ( ) ( ) ( )
10)Show that the poles of tangents to the circle
x2+y2=r2 w.r.to the circle (x + a) 2+y2=2a2 lie on
y2+4ax=0.
Sol: Let P(x1, y1) be the pole of tangents of the circle
x2+y2=r2 w. r. to the circle S(x + a) 2+y2=2a2x 2+y2 +2ax-a2=0Now the polar of p w.r.t to S=0 is S1=0
xx1+yy1+a(x+x1) - a2=0x(x1+a) +yy1+ (ax1- a2) =0.. (1)
()
( ) ( ) *( ) +
*( ) *( ) + ( ) ( ) The equation of locus of P(x1, y1) is y2+4ax=0
11)If are the angles of inclination of tangentsthrough a point p to the circle x2+y2=r2, then find
locus of p when cot =k.Sol: given equation of the circle x2+y2=r2. (1): Let P(x1, y1) be the any point on the locus
Equation of tangent through p with slope m is
Y= This passes through P(x1, y1)
= - S.OB ( ) ( ) ( )
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)( ) () ( ) Where m1, m2 be the slopes of the tangents which
make angles m1=tan m2=tanGiven cot t =k. +
+
+
( ) The equation of locus of P(x1, y1) is( )
12)If the chord of contact of a point P with respect tothe circle x2+y2=a2 cut the circle at A and B such
that AOB =900 then show that P lies on the circle
x2+y2=2a2.
Sol: Given circle x2+y2=a2... (1)Let p(x1, y1) be a point and let the chord of contact of
it cut the circle in A and B such that AOB=900. The
equation of the chord of contact of p(x1, y1) with
respect to (1) sis xx1+yy1-a2=0.. (2)The equation to the pair of lines OA and OB is given
by x2+y2-a2. /
Or ( ) ( ) ( ) ( ) ()Since AOB=900, we have the coefficient of () () Hence the point p(x1, y1) lies on the circle x2+y2=2a2.
13)Prove that the combined equation of the pair oftangents drawn from an external point p ( ) tothe circle S=0 is .Sol: given that p ( ) is an external point on S=0let AB be the chord of contact of P to the circle S=0
and its equation is Let Q ( ) be any point on the locusi.e. , be a point on the pair of tangents
the ratio that the line AB divides PQ can be
determined in 2ways :
(1) PB:QB is equal to : ()(2) The line
Points p ( ), Q ( ) in the ratio- ()From (1), (2) we get
Hence the equation of the locus of Q ( ) is .* ( ) replaced by (x, y)}
14)Show that the area of the triangle formed by thetwo tangents through p(x1, y1) to the circles
S x2+y2+2gx+2fy+c=0 and the chord of contact ofp with respect to S=0 is
()
, where r is the radiusof the circle.
SAQ:1. Find the equations of the tangents to the circle
x2+y 2-4x+6y-12=0 which are parallel to Sol: given equation of the circle x2+y2-4x+6y-12=0
Centre (2, -3) and radius (r)
=() () = The given line x+y- ()Any line parallel to the above line is
()If (2) touches the given circle then r = distancefrom (2, -3)
()()
(k-1) = k
H rquird q f tgts r 2. Find the equation of the tangent to x2+y 2-2x+4y=0 at
(3, -1). Also find the equation of tangent parallel to it.Sol: given equation of the circle
x2+y2- ()Centre (1, -2) and radius (r)=() () =The equation of tangent at (3, -1) is
x(3)+y(-1)-1(x+3)+2(y-1)=0
() () Rquird q f th tgt t () d it is prlllto (2) is ( ) ( ) ( ) ( ) ()
( ) ( )
( ) ( ) 3. Find the equations of the tangents to the circle
x2+y 2+2x-2y-3=0 which are perpendicular to Sol: given equation of the circlex2+y2+2x-2y- ()Centre (-1, 1) and radius (r)
=() () =The given ()Any line to the above line is ( ) Since (3) is tangent to (1) r = distance from (-1, 1)()()
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k H rquird q f tgts r
4. Show that the line 5x+12y-4=0 touches the circlex2+y 2-6x+4y+12=0, also find the point of contact.Sol: given equation of the line 5x+12y-4=0 andequation of the circle x2+y2-6x+4y+12=0
Centre (3, -2) and radius (r)
=() () = If given line touches the given circle then radius of
circle = distance from centre (3, -2) to given lined()()()()
r d th giv stright li tuhs th giv irl 5. Show that the tangent at (-1, 2)of the circle
x2+y 2-4x-8y+7=0 touches the circle x2+y 2+4x+6y=0and also find its point of contact.Sol: equation of the tangent at (-1, 2) to the circle (1)() ( ) ( ) ()For the circle x2+y 2+4x+6y=0 centre (-2, -3), r=() () =Distance from centre (-2, -3) to given line (1)
()()()()
so the line(1) also touches the 2nd circle.
( ) ()
( )
(() () )
()
Coordinate of point of contact =(1, -1.)
6. Find the equations of normal to the circle x2+y 2-4x+6y+11= 0 at (3, 2) also find the other point wherenormal meets the circleSol: given equation of the circle
x2+y2-4x- ()Centre C (2, 3) = (-g, -f)
Given point A (3, 2) = ( )The equation of the normal is
( )( ) ( )( ) ( ) ( ) ( ) ( ) centre ofthe circle is mid point of A and B
0 1 ( )
( ) ()
7. Find the area of the triangle formed by the no rmal at(3, -4) to the circle x2+y 2+22x-4y+25=0 with thecoordinate axes.Sol: given equation of the circle
x2+y2-22x- ()Centre C (11, 2) = (-g, -f)
Given point A (3, -4) = ( )The equation of the normal is
( )( ) ( )( ) ( ) ( ) ( ) ( ) Area of the triangle formed by the normal with the
coordinate axes = |
| |
()
() |
8. Find the pole of 3x+4y-45=0 w ith respect tox2+y 2-6x-8y+5=0sol: given equation of the circlex2+y2-6x- ()Centre (3, 4) and r =() () =Given line 3x+4y-45=0 here l=3, m=4 and n=-45
The pole =.
/=. ()()() ()()()/=
. ()
()
/
( )=(6. 8)9. If a point P is moving such that the lengths of tangentsfrom P to the circle x2+y 2-6x-4y-12=0 andx2+y 2+6x+18y+26= 0 are in the ratio 2:3 the find theequation of the locus of p.Sol: let P(x1, y1) be any point on the locus and PT1,PT2 be the lengths of tangents from p to the given
circles, and then we have
S.O.B
( )( )
( )
=0 th quti f lus f p is 5x2+5y2-78x-108y-212=0
10.Find the length of chord intercepted by the circlex2+y 2-8x-2y-8=0 on the line x+y+1=0.Sol: given equation of the circlex2+y2-8x-2y- ()Centre (4, 1) and r =() () =
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Given line x+y+1=0
Distance from centre (-2, -3) to given line (1) ()()()()
length of chord intercepted by the circle is
2 = 2 =211.Find the eq uation of the circle with centre (-2, 3)
cutting a chord length 2 units on 3x+4y+4= 0.Sol: given centre C (-2, 3)
Given equation of the chord ( )d= Distance from centre C (-2, 3) to given line (1)d =
()()()()
Given length of chord 2 = 2 = 1 (d=2)
Rquird q f th irl is( ) ( ) ( ) ( ) x2+y2+4x-6y+8=0
12.Find the equation of pair of tangents drawn from (0,0) to x2+y 2+10x+10y+40=0Sol: given equation of the circle
x2+y2+10x+ ()P(x1, y1= (0, 0)
() () ( ) ( )
02+02+10(0) +10(0) +40=40 ( ) ( x2+y2+10x+10y+40)()25( ) ( x2+y2+10x+10y+40)()* +
* +*
13.Find the value of k if kx+3y-1=0, 2x+y+5=0 areconjugate lines with respect to the circlex2+y 2-2x-4y-4=0.
Sol: condition is ( ) ( )( )
( ) (() () )(() () )
(k) (-k-5) (-9)k k k - KVSAQ
1 Find centre and radius of circles given by 2 2 21 2 2 0m x y cx mcy
Sol:Given equation of the circle
( )
Centre (-g, -f) = (
)
And r= ( ) (
)
=
=()() =c
2 Find centre and radius of x2+y2+6x+8y-96=0.Sol: Given equation of the circle is x2+y2+6x+8y-96=0.
Compare with x2+y2+2gx+2fy+c=0.
Centre (-g, -f) = (-3, -4)
Radius r= =() () =
3 Find the length of the tangent from (3, 3) to thecircle x2+y2+6x+8y+26=0.
Sol: Given equation of the circle isx2+y2+6x+8y+26=0.
Length of the tangent from (3, 3)= () () = =
4 Find the power of the point (3, 4) w. r. t the circlex2+y2-4x-6y-12=0.
Sol: Given equation of the circle is
x2+y2-4x-6y-12=0.
Power of the point is ()()=9+16-12-24-12=-23.
5 Find centre and radius of 3x2+3y2-6x+4y-4=0.Sol: Given equation of the circle is
3x2+3y2-6x+4y-4=0.
. / ( )
=
.
6 If length of tangent from (2,5) to the circle is, then find k.Sol: Length of the tangent from (2, 5)= () ()
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S.O.B
k=-2
7 Obtain parametric equation of the circle Ifx2+y2-6x+4y-12=0, x2+y2+6x+8y-96=0.
Sol: centre (3, -2), c=-12And r= () =
Parametric equation of the circle
X= s , s i s si
8 If x2+y2-4x+6y+c=0 represents a circle withradius 6, find the value of c.
Sol: centre (2, -3) and
r= () = S.O.B13+c=36
=23.9 If x2+y2-4x+6y+a=0 represents a circle with
radius 4, find the value of a.
Sol:
10 If x2+y2+6-8y+c=0 represents a circle with radius6, find the value of c.
Sol:
11 If x2+y2+ax+by-12=0 is a circle with centre (2, 3),find the value of a, b and radius.
Sol: x2+y2+2ax+2by-12=0
Since centre (-, -
) = (2, 3)
a=-4 and b=-6Radius
r= () = =
12 If 3x2+2hxy+by2-5x+2y-3=0 represents a circle,find a, b.
Sol: If ax2+2hxy+by2+2gx+2fy+c=0 represents a circle
then a=b and h=0
b=3 and h=0
13 Find the equation of the circle with (1, 2), (4, 5) asends of a diameter.
Sol: the equation of the circle with ( ), ( ), asends of a diameter.
Is (x-) (x-) +(y-) (y-) =0(x-1)(x-4)+(y-2)(y-5)=0
14 Find the equation of circle passing through (5, 6)and having centre (-1, 2).
Sol: given centre C (-1, 2), point on the circle P(5, 6)
Radius (r) =CP=
( )
( )=
equation of the circle with centre (-1, 2) and radius
is ( ) ( ) ()x2+y2+2x-4y+5-52=0.
x2+y2+2x-4y-47=0.
15 Find the equation of circle passing through (0, 0)and having centre (-4, -3).
Sol: given centre C (-4, -3), point on the circle P(0, 0)
Radius (r) =CP=() ( )=
equation of the circle with centre (-4, -3) and radius is ( ) ( ) ()x2+y2+8x+6y+16+9-25=0.x2+y2+8x+6y=0.
16 Find the equation of the circle passing through (2,3) and concentric with
01512822
yxyx .
Sol: equation of the circle concentric with
x2+y2+8x+12y+15=0, is in the form
x2
+y2
+8x+12y+k=0.Since it is passes through (2, 3)
4+9+16+36+k=0 65+ k=0 k=-65 required eqn of the circle is x2+y2+8x+12y-65=0,
17 Find the pole of x + y + 2 = 0 with respect to thecircle x2 + y2 4x + 6y 12 = 0.
Sol:
18 Show that the points (4, -2), (3, -6) are conjugatew.r.to the circle x2+y2=24.
Sol: ( ) ( )
(4) (3) + (-2) (-6)-2412+12 -24=0
19 If (4, k), (2, 3) are conjugate points with respect
to the circle x2 + y2 = 17 then find k.
Sol: ( ) ( )
(4) (2) + (k) (3) =17 8+ 3k =17 3k = 17 8 3k = 9 k=3.
20.Show that the line
Sol: the straight line
x2+y2+2gx+2fy+c=0.
()
() ()
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