Aims Tutorial 2B Important Questions

7/28/2019 Aims Tutorial 2B Important Questions

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Aims tutorial 2B important questions & AnsLAQ (2 )

1) Find the equation of the equation of the circlepassing 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.(1)

(2, -1) lies on S=0

(2)2+ (-1)2+2g (2) +2f (-1) +c=04g - 2f +c =-5..(2)(3, 2) lies on S=0

(3)2+ (2)2+2g (3) +2f (2) +c=06g + 4f +c =-13.(3)

Solving eqn (1) & (2) (2) & (3)

2g + 2f +c =-2 4g - 2f +c =-5

4g - 2f +c =-5 6g + 4f +c =-13

-2g+4f =3(4) -2g-6f=8.(5)

Solving eqn (4) & (5) sub f value in eqn (4)

-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 (1)

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)}


x2+y2+2gx+2fy+c=0. (*)

A (1, 1) lies on S=0 (1)2+ (1)2+2g (1) +2f (1) +c=02g + 2f +c =-2..(1)

B (-6, 0) lies on S=0

(-6)2+ (0)2+2g (-6) +2f (0) +c=0-12g +c =-36..(2)C (-2, 2) lies on S=0

(-2)2+ (2)2+2g (-2) +2f (2) +c=0-4g + 4f +c = - 8.(3)

Solving eqn (1) & (2) (2) & (3)

2g + 2f +c =-2 -12g +0 +c =-36

-12g +c =-36 -4g + 4f +c =-8

14g+2f =34 (4) -8g-4f=-28. (5)

Solving eqn (4) & (5) sub f value in eqn (4)

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 (1)

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)}


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.. (1)

B (0, 1) lies on S=0

(0)2+ (1)2+2g (0) +2f (1) +c=0 2f +k =-1.. (2)

C (4, 5) lies on S=0

(4)2+ (5)2+2g (4) +2f (5) +c=08g + 10f +k =-41. (3)


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Aims tutorial 2B important questions & AnsSolving eqn (1) & (2) (2) & (3)

4g + 0 + k =-4 0 + 2f + k =-1

0 + 2f +k =-1 8g + 10f +k =-41

4g - 2f =-3 (4) -8g-8f=40. (5)

Solving eqn (4) & (5) sub f value in eqn (4)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 (1)

4g +k=-44 (- ) +k=-4

the required eqn of the circle isx2+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) =0c=1, c=

4) Find the equation of the equation of the circlepassing through the points (4, 1), (6, 5) and whose

centre lies on {H/W (2, -3), (-4, 5) and 4x+3y+1=0, (4, 1), (6, 5), and

4x+y-16=0}


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..(1)

(6, 5) lies on S=0

(6)2+ (5)2+2g (6) +2f (5) +c=012g +10f+c =-61.. (2)Solving eqn (1) & (2)

8g + 2f +c =-17

12g +10f+c =-61

-4g-8f = 44.(3)

Given centre (-g, -f) lies on 4x+3y-16=0

4 (-g) +3(-f)-24=0

4g+3f+24=0 (4)Solving eqn (3) & (4)

-4g-8f = 44

4g+3f=-24

-5f = 20 f=-4From eqn (4)4g+3f+24=04g+3(-4)=-24

4g=-24+124g=-12g=-3From (1)

8g + 2f +c =-17

8(-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 lies on

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.. (1)

(4, 5) lies on S=0

(4)2+ (5)2+2g (4) +2f (5) +c=08g +10f+c =-41.. (2)Solving eqn (1) & (2)

-4g + 6f +c =-13

8g +10f+c =-41

-12g-4f = 28.(3)

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 isx2+y2+2( )x+2(0)y+( )=0.3 x2+3y2-14x-67=0.


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Aims tutorial 2B important questions & Ans6) Show that the circles touch each other. Also find

the 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= r1=() ()

() () () ( ) =5 given two circles touch externallyThe point of contact P divides 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=0Or 4x-3y+6=0.

7) Find the equation of the pair of transverse commontangents 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

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 =0x2+y2-4x-6y-12=0, x2+y2+6x+18y+26=0.(H/W)


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Aims tutorial 2B important questions & Ans8) Find the equation of direct common tangents to the

circles x2+y2+22x-4y-100=0,

x2+y2-22x+4y+100=0.

Sol: given equation of the circles

S

x2+y2+22x-4y-100=0, and

Centre C1= (-11, 2) C2 (11,-2)

And r= r1=() ()

() ()

() ( ) = There exist two direct common tangents

External centre of similitude Q divides

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 =0Method II

Let the eqn of tangent through Q (22, -4) with slope

m is (Y-)=m(x-)(y+4) =m(x-22)mx-y-(22m+4)=0. (1)

The perpendicular distance from () ()

()()()()

()

|11m+2|=5 {S.O.B} ()=25( )

121 96 96 24m (4m+3)-7(4m+3) =0

(24m-7)(4m+3)=0m= , m=

the equation of the tangent is (y+4) = ( )(x-22)24(y+4) =7(x-22)24y+96= 7x-1547x-24y-250=0And is (y+4) = (

)(x-22)

4(y+4) =-3(x-22)4y+16= -3x+663x+4y-50=0


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Aims tutorial 2B important questions & Ans9) Find the equation of the circles which touch

2x-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 circlex2+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 findlocus 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 ( ) ( ) ( ) )( ) () ( ) Where m1, m2 be the slopes of the tangents which

make angles m1=tan m2=tanGiven cot =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 ( ) ( ) ( ) ( ) (3)Since AOB=900, we have the coefficient of

() () Hence the point p(x1, y1) lies on the circle x2+y2=2a2.


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Aims tutorial 2B important questions & Ans13)Prove that the combined equation of the pair of

tangents 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 : .(1)(2) The line

Points p ( ), Q ( ) in the ratio-. (2)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+y2-4x+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-8=0.. (1)

Any line parallel to the above line is

()If (2) touches the given circle then r = distancefrom (2, -3)()() (k-1) = k=1

Hence required eqn of tangents are x+y+1

2. Find the equation of the tangent to x2+y2-2x+4y=0at (3, -1). Also find the equation of tangent parallel

to it.

Sol: given equation of the circle

x2+y2-2x+4y=0 ..(1)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

. (2) () Required eqn of the tangent to (1) and it is

parallel to (2) is ( ) ( ) ( ) ( ) () ( ) ( ) ( ) ( )

3. Find the equations of the tangents to the circlex2+y2+2x-2y-3=0 which are perpendicular to


x2+y2+2x-2y-3=0 ..(1)

Centre (-1, 1) and radius (r)

=() () =The given 0.. (2)Any line to the above line is ( )Since (3) is tangent to (1) r = distance from (-1, 1)()()

k = Hence required eqn of tangents are

4. Show that the line 5x+12y-4=0 touches the circlex

2

+y2

-6x+4y+12=0, also find the point of contact.Sol: given equation of the line 5x+12y-4=0 and

equation 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, the given straight line touches the givencircle.

5. Show that the tangent at (-1, 2)of the circlex2+y2-4x-8y+7=0 touches the circle x2+y2+4x+6y=0

and also find its point of contact.

Sol: equation of the tangent at (-1, 2) to the circle

(1)() ( ) ( ) . (1)For the circlex2+y2+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.

( )


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Aims tutorial 2B important questions & Ans ()

( )

(() () )

()

Coordinate of point of contact =(1, -1.)

6. Find the equations of normal to the circle x2+y2-4x+6y+11=0 at (3, 2) also find the other point

where normal meets the circleSol: given equation of the circle

x2+y2-4x-6y+11=0 ..(1)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 normalat (3, -4) to the circle x2+y2+22x-4y+25=0 with the

coordinate axes.


x2+y2-22x-4y+25=0 ..(1)

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 with respect tox2+y2-6x-8y+5=0

sol: given equation of the circle

x2+y2-6x-8y+5=0 ..(1)

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 oftangents from P to the circle x2+y2-6x-4y-12=0 and

x2+y2+6x+18y+26=0 are in the ratio 2:3 the find the

equation 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

3

S.O.B ( )

( )( )

=0 the equation of locus of p is5x2+5y2-78x-108y-212=0

10.Find the length of chord intercepted by the circlex2+y2-8x-2y-8=0 on the line x+y+1=0.


x2+y2-8x-2y-8=0 ..(1)

Centre (4, 1) and r =() () = 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 equation 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) Required eqn of the circle is

( ) ( )

( )

( )

x2+y2+4x-6y+8=0


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Aims tutorial 2B important questions & Ans12.Find the equation of pair of tangents drawn from

(0, 0) to x2+y2+10x+10y+40=0


x2+y2+10x+10y+40=0 ..(1)

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 circle

x2+y2-2x-4y-4=0.

Sol: condition is ( ) ( )( )

( ) (() () )(() () )

9 (2k+3) = (-k-5) (-9)2k + 3 = k+ 5 k = 5- 2 K=2.

VSAQ

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 is

x2+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 circlex

2

+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)= () ()

S.O.B

k=-27 Obtain parametric equation of the circle If

x2+y2-6x+4y-12=0, x2+y2+6x+8y-96=0.

Sol: centre (3, -2), c=-12

And r= () = Parametric equation of the circle

X= ,

8 If x2+y2-4x+6y+c=0 represents a circle with radius6, find the value of c.


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Aims tutorial 2B important questions & AnsSol: centre (2, -3) and

r= () = S.O.B13+c=36=23.

9 If x2+y2-4x+6y+a=0 represents a circle with radius4, find the value of a.Sol:

10 If x2+y2+6-8y+c=0 represents a circle with radius 6,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=-6Radiusr= () = =

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=013 Find the equation of the circle with (1, 2), (4, 5) as

ends of a diameter.

Sol: the equation of the circle with ( ), ( ),as ends 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) andradius 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) andradius 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=-65required 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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Aims Tutorial 2B Important Questions