2b Maths Important (2)

Aims tutorial 2B important questions & Ans

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=0 2g + 2f +c =-2………..⇒ ①s (2, -1) lies on S=0 (2)⇒ 2+ (-1)2+2g (2) +2f (-1) +c=0 4g - 2f +c =-5………..⇒ ②

(3, 2) lies on S=0 (3)⇒ 2+ (2)2+2g (3) +2f (2) +c=0 6g + 4f +c =-13……….⇒ ③

Solving eq’’n & ① ② & ② ③2g + 2f +c =-2 4g - 2f +c =-54g - 2f +c =-5 6g + 4f +c =-13-2g+4f =3…………④ -2g-6f=8……….…⑤

Solving eq’’n & ④ ⑤ sub f value in eq’’n ④

-2g+4f =3 -2g+4(−12

) =3

-2g-6f=8 -2g=3+2

10f=-5 g=⇒−52

f=-⇒12

Sub the value of g and f in eq’’n ①

2 (⇒−52

) + 2(-12

) +c =-2

-5-1+c=-2 c=4⇒ ⇒

the required eq’’n of the circle is ∴x2+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)}


A (1, 1) lies on S=0 (1)⇒ 2+ (1)2+2g (1) +2f (1) +c=0 2g + 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 eq’’n & ① ② & ② ③2g + 2f +c =-2 -12g +0 +c =-36-12g +c =-36 -4g + 4f +c =-814g+2f =34…………④ -8g-4f=-28…….…⑤


14g+2f =34 14g+2f =34 4g+2f=14 14(2) +2f =34⇒

10g=20 2f=⇒ 34-28

g=2 f=3⇒

Sub the value of g and f in eq’’n ①

2 (⇒ 2) + 2(3) +c =-2

4+6+c=-2 c=-12⇒ ⇒

the required eq’’n of the circle is ∴x2+y2+4x+6y-12=0.Now substituting D (-2, -8) in the above eq’’n, 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, and then find the value of c.{H/w (1, 2), (3, -4), (5, -6), (c, 8)}


A (2, 0) lies on S=0 (2)⇒ 2+ (0)2+2g (2) +2f (0) +c=0 4g +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=0 8g + 10f +k =-41……….⇒ ③

Solving eq’’n & ① ② & ② ③4g + 0 + k =-4 0 + 2f + k =-10 + 2f +k =-1 8g + 10f +k =-414g - 2f =-3…………④ -8g-8f=40……….…⑤


4g-2f =-3 4g-2(−176

) =-3

-4g-4f=20 4g=-3-173

-6f=17 g=⇒−2612

f=-⇒176

g=-⇒136

Sub the value of g and f in eq’’n ①4g +k=-4⇒

4 (-⇒136

) +k=-4

⇒ k=−4+ 263⇒ k=14

3

the required eq’’n of the circle is ∴

x2+y2+2(−136

)x+2(−176

)y+143

=0.


⇒ 3x2+3y2-13x-17y+14=0 given 4 points are concyclic, ∴

D (0, c) lies on the above circle3(0)⇒ 2+3(c)2-13(0)-17(c)+14=03c⇒ 2-17c+14=03c⇒ 2-3c-14c+14=03c(c-1)-14(c-1) =0⇒(c-1) (3c-14) =0⇒(c-1) =0 or (3c-14) =0⇒

c=1, c=∴143

4) Find the equation of the equation of the circle passing through the points (4, 1), (6, 5) and whose centre lies on

4 x+3 y−24=0 .{H/W (2, -3), (-4, 5) and 4x+3y+1=0, (4, 1), (6, 5), and 4x+y-16=0}


(4, 1) lies on S=0 (4)⇒ 2+ (1)2+2g (4) +2f (1) +c=0 8g + 2f +c =-17………..⇒ ①

(6, 5) lies on S=0 (6)⇒ 2+ (5)2+2g (6) +2f (5) +c=0 12g +10f+c =-61………..⇒ ②

Solving eq’’n & ① ② 8g + 2f +c =-1712g +10f+c =-61

-4g-8f = 44……….③Given centre (-g, -f) lies on 4x+3y-16=0

4 (-g) +3(-f)-24=0⇒4g+3f+24=0…………⇒ ④

Solving eq’’n & ③ ④

-4g-8f = 444g+3f=-24 -5f = 20 f=-4⇒From eq’’n ④

4g+3f+24=0 ⇒4g+3(-4)=-24⇒

4g=-24+12⇒4g=-12 g=-3⇒From (1)8g + 2f +c =-17

8(-3)+2(-4)+c=-17⇒c=-17+24+8⇒

C=15Required eq’’n of the circle is x∴ 2+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=0 8g +10f+c =-41………..⇒ ②

Solving eq’’n & ① ② -4g + 6f +c =-13 8g +10f+c =-41 -12g-4f = 28……….③Given centre (-g, -f) lies on X- axis

-f=0 or f=0⇒-12g-0=28

g=-⇒2812

=−73.

Substituting f=0,g=-7/3 in eq’’n (1)-4g + 6f +c =-13

-4(73.) +6(0) +c=-13

C=-13-283

C= -673.

Required eq’’n of the circle is x∴ 2+y2+2(−73

)x+2(0)y+(−673

)=0.

3 x⇒ 2+3y2-14x-67=0.

6) 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

S≡ x2+y2-6x-2y+1=0, and

S ’≡x 2+ y 2+2x−8 y+13=0

Centre C1= (3, 1) C2 (-1, 4)

And r = √ g2+f 2−Cr⇒ 1=√ (3 )2+(1 )2−1=3

r2=√(1)2+(4)2−13=2

r1+r2=3+2=5

C1C2=√(−1−3)2+(4−1)2=√16+9=5

r1+r2=C1C2

given two circles touch externally∴

The point of contact P divides C1C2∈the ratio=r1: r2=3 :2

P=[m x2+nx1m+n,m y2+n y1m+n ]=

[ 3 (−1)+2(3)3+2,3 (4 )+2(1)3+2 ]

=[−3+65 ,12+25 ]=[ 35 , 145 ]

Equation of common tangent at point of contact is the radical axis of the circles S-S’=0

(x2+y2-6x-2y+1=0)- ( x2+ y 2+2 x−8 y+13) =0

-8x+6y-12=0⇒


Or 4x-3y+6=0.

a) x2+y2-4x-6y-12=0, x2+y2+6x+18y+26=0.(H/W)7) 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

S ’≡x2+ y2−22 x+4 y+100=0

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

And r = √ g2+f 2−Cr⇒ 1=√ (11)2+(−2 )2+100=√121+4+100=√225=15

r2=√(−11)2+(2)2−100=√121+4−100=√25=5

r1+r2=15+5=20

C1C2=√(−11−11)2+(2+2)2=√484+16=√500

¿C1C2∨¿ r1+r2

There exist two direct common tangents

External centre of similitude ‘Q’ divides

C1C2∈the ratio=−r1 :r2=−15 :5=−3:1

Q=[m x2−n x1m+n,m y2−n y1m+n ]=

[3 (11 )−1(−11)3−1

,3 (−2 )−1(2)3−1 ]

=[ 33+112 ,−6−22 ]=[ 442 ,−82 ]=[22 ,−4]

The equation to the pair of direct common tangents through Q(22, -4)

to the circle S≡x2+ y2+22x−4 y−100=0 is S12=S S11

S1=x x1+ y y1+g (x+x1 )+ f ( y+ y1 )+c=0

S11=x12+ y1

2+2g x1+2 f y1+c=0 Where (x1 , y1)=(22, -

4)

⇒S12=S S11

⇒[ x (22 )+ y (−4 )+11 ( x+22 )−2 ( y−4 )−100 ]2

=

[ x2+ y2+22x−4 y−100 ] [(22)2+(−4 )2+22(22)−4 (−4)−100 ]

⇒[22x−4 y+11 x+242−2 y+8−100 ]2

=

[ x2+ y2+22x−4 y−100 ] [484+16+484+16−100 ]

⇒[33 x−6 y+150 ]2=[ x2+ y2+22x−4 y−100 ] [900 ]

⇒9 [11 x−2 y+50 ]2=[ x2+ y2+22x−4 y−100 ] [900 ]

⇒[11 x−2 y+50 ]2=[ x2+ y2+22x−4 y−100 ] [100 ]

{121⇒ x2+4 y2+2500−44 xy−200 y+1100 x}

=[100 x2+100 y2+2200 x−4 y 00−10000 ]

21⇒ x2−96 y2−44 xy−200 y+12500=0

Method II

Let the eq’’n of tangent through Q (22, -4) with slope ‘m’ is

(Y-y1)=m(x-x1)

(y+4) =m(x-22)⇒

mx-y-(22m+4)=0………………. (1)

The perpendicular distance from C2 (11 ,−2 )¿ (1 ) is r2=5

⇒¿a x1+b y1+c∨¿

√a2+b2=5¿

⇒¿m (11 )−1 (−2 )−22m−4∨ ¿√(m)2+(−1)2

=5¿

⇒|11m−2−22m−4|

√m2+12=5⇒

|−11m−2|√(m )2+1

=5

|11m+2|=5√m2+1 {S.O.B} ⇒(11m+2)2=25(m2+1)

121⇒ m2+2+44m=25m2+25

96⇒ m2+44m−21=0

96⇒ m2+72m−28m−21=0

24m (4m+3)-7(4m+3) =0⇒

(24m-7)(4m+3)=0⇒

m=⇒724

, m=−34

the equation of the tangent is (y+4) = (∴724

)(x-22)

24(y+4) =7(x-22)⇒


24y+96= 7x-154 7x-24y-250=0⇒ ⇒

And is (y+4) = (−34

)(x-22)

4(y+4) =-3(x-22)⇒

4y+16= -3x+66 3x+4y-50=0⇒ ⇒

8) Find the equation of the pair of transverse common tangents to the circles

x2+y2-4x-10y+28=0, x2+y2+4x-6y+4=0.

Sol: Sol: given equation of the circles

S≡ x2+y2-4x-10y+28=0, and

S ’≡x2+ y2+4 x−6 y+4=0

Centre C1= (2, 5) C2 (-2, 3)

And r = √ g2+f 2−Cr⇒ 1=√ (−2 )2+(−5 )2−28=√4+25−28=√1=1

r2=√(2)2+(−3)2−4=√4+9−4=√9=3

r1+r2=1+3=4

C1C2=√(−2−2)2+(3−5)2=√16+4=√20

¿C1C2∨¿ r1+r2

There exist two transverse common tangents

Enternal centre of similitude ‘P’ divides

C1C2∈the ratio=r1: r2=1 :3

Q=[m x2−n x1m+n,m y2−n y1m+n ]=

[ 3 (11 )−1(−11)3−1

,3 (−2 )−1(2)

3−1 ]=[ 33+112 ,

−6−22 ]=[ 442 ,−82 ]=[22 ,−4]

The equation to the pair of direct common tangents through Q(22, -4)

to the circle S≡x2+ y2+22x−4 y−100=0 is S12=S S11

S1=x x1+ y y1+g (x+x1 )+ f ( y+ y1 )+c=0

S11=x12+ y1

2+2g x1+2 f y1+c=0 Where (x1 , y1)=(22, -

4)

⇒S12=S S11

⇒[ x (22 )+ y (−4 )+11 ( x+22 )−2 ( y−4 )−100 ]2

=

[ x2+ y2+22x−4 y−100 ] [(22)2+(−4 )2+22(22)−4 (−4)−100 ]

⇒[22x−4 y+11 x+242−2 y+8−100 ]2

=

[ x2+ y2+22x−4 y−100 ] [484+16+484+16−100 ]

⇒[33 x−6 y+150 ]2=[ x2+ y2+22x−4 y−100 ] [900 ]

⇒9 [11 x−2 y+50 ]2=[ x2+ y2+22x−4 y−100 ] [900 ]

⇒[11 x−2 y+50 ]2=[ x2+ y2+22x−4 y−100 ] [100 ]

{121⇒ x2+4 y2+2500−44 xy−200 y+1100 x}

=[100 x2+100 y2+2200 x−4 y 00−10000 ]

21⇒ x2−96 y2−44 xy−200 y+12500=0

Method II

Let the eq’’n of tangent through Q (22, -4) with slope ‘m’ is

(Y-y1)=m(x-x1)

(y+4) =m(x-22)⇒

mx-y-(22m+4)=0………………. (1)

The perpendicular distance from C2 (11 ,−2 )¿ (1 ) is r2=5

⇒¿a x1+b y1+c∨¿

√a2+b2=5¿

⇒¿m (11 )−1 (−2 )−22m−4∨ ¿√(m)2+(−1)2

=5¿

⇒|11m−2−22m−4|

√m2+12=5⇒

|−11m−2|√(m )2+1

=5

|11m+2|=5√m2+1 {S.O.B} ⇒(11m+2)2=25(m2+1)

121⇒ m2+2+44m=25m2+25

96⇒ m2+44m−21=0

96⇒ m2+72m−28m−21=0

24m (4m+3)-7(4m+3) =0⇒

(24m-7)(4m+3)=0⇒


m=⇒724

, m=−34

the equation of the tangent is (y+4) = (∴724

)(x-22)

24(y+4) =7(x-22)⇒

24y+96= 7x-154 7x-24y-250=0⇒ ⇒

And is (y+4) = (−34

)(x-22)

4(y+4) =-3(x-22)⇒

4y+16= -3x+66 3x+4y-50=0⇒ ⇒

9) Find the equation of the circles which touch

2x-3y+1=0 at (1, 1) and having radius √13.

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.

11) If θ1 , θ2are the angles of inclination of tangents through a

point p to the circle x2+y2=r2, then find locus of p when cot

θ1 cot θ2 =k.

12) Show that the area of the triangle formed by the two tangents through p(x1, y1) to the circles

S≡ x2+y2+2gx+2fy+c=0 and the chord of contact of p with

respect to S=0 is r (S11)32

S11+r2

, where r is the radius of the circle.

13)

12 Find centre and radius of circles given by

Sol: Given equation of the circle

√1+m2 (x2+ y2 )−2cx−2mcy=0

⇒ x2+ y2− 2cx

√1+m2− 2mcy

√1+m2=0

Centre (-g, -f) = (c

√1+m2,mc

√1+m2)

And r=√ g2+f 2−c=√(c

√1+m2)2

+(mc

√1+m2)2

−0

=√ c2

1+m2+ m

2 c2

1+m2=√c2 (1+m¿¿2)

(1+m2)¿=c

3 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=√ g2+f 2−c=√(−3)2+(−4)2+96

=√121=11.

4 Find the length of the tangent from (3, 3) to the circle x2+y2+6x+8y+26=0.Sol: Given equation of the circle is x2+y2+6x+8y+26=0.

Length of the tangent √s11 from (3, 3)

=√32+32+6 (3 )+8 (3 )+26=√9+9+18+24+26=√86

5 Find the power of the point (3, 4) w. r. t the circle x2+y2-4x-6y-12=0.Sol: Given equation of the circle is x2+y2-4x-6y-12=0.

Power of the point is s11=32+42−4 (3)−6 (4)−12=9+16-12-24-12=-23.

6 Find centre and radius of 3x2+3y2-6x+4y-4=0.Sol: Given equation of the circle is 3x2+3y2-6x+4y-4=0.

⇒x2+ y2−6x3

+ 4 y3

−43=0

⇒ centre(1,−23 )∧c=−43

r=√g2+ f 2−c=√12+(−23

)2

+ 43

¿√ 9+4+129=√ 259 =5

3.

7 If length of tangent from (2,5) to the circle

x2+ y2−5 x+4 y+k=0 is√ 37, then find k.

Sol: Length of the tangent √s11 from (2, 5)

=√22+52−5 (2 )+4 (5 )+k=√37⇒√4+25−10+20+k=√37

S.O.B

39+k=37⇒ k=37−39k=-2∴

8 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=√32+(−2)2+12=√9+4+12=5 Parametric equation of the circle


X=x1+r cosθ, y= y1+r sinθX=3+5cosθ ,Y=−2+5sin θ

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

Sol: centre (2, -3) and r=√22+(−3)2+c=√4+9+c=6

S.O.B13+c=36⇒=23.∴

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

11 If x2+y2+6-8y+c=0 represents a circle with radius 6, find the value of c.Sol:

12 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 (-a2

, -b2

) = (2, 3)

a=-4 and b=-6⇒

Radius r=√ g2+f 2−c=√22+(3)2+12=√4+9+12=√25=5.

13 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∴

14 Find the equation of the circle with (1, 2), (4, 5) as ends of a diameter.

Sol: the equation of the circle with (x1 , y1), (x2 , y2), as ends of a

diameter.

Is (x-x1) (x-x2) +(y-y1) (y-y2) =0

(x-1)(x-4)+(y-2)(y-5)=0⇒

⇒x2+ y25 x−7 y+14=015 Find the equation of circle passing through (5, 6) and having

centre (-1, 2).Sol:

16 Find the equation of circle passing through (0, 0) and having centre (-1, 2).Sol:

17 Find the equation of the circle passing through (2, 3) and

concentric withx2+ y2+8 x+12 y+15=0 .

Sol:18 Find the pole of x + y + 2 = 0 with respect to the circle x2 + y2 – 4x +

6y – 12 = 0.Sol:

19 Show that the points (4, -2), (3, -6) are conjugate w.r.to the circle x2+y2=24.Sol:

20 If (4, k), (2, 3) are conjugate points with respect to the circle x2 + y2 = 17 then find k.Sol:

21 Show that the lines 2 x+3 y+11=0 and

2 x−2 y−1=0 are conjugate with respect to the circle

x2+ y2+4 x+6 y+12=0 .

22

2b Maths Important (2)

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