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………..①
s (2, -1) lies on S=0 ⇒ (2)2+ (-1)2+2g (2) +2f (-1) +c=0 ⇒4g - 2f +c =-5………..②
Sub the value of g and f in eq’’n ① ⇒2 ( ) + 2( ) +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)
(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 2g + 2f +c =-2 4g - 2f +c =-5 -2g+4f =3…………④
②&③ 4g - 2f +c =-5 6g + 4f +c =-13 -2g-6f=8……….…⑤
Solving eq’’n ④ & ⑤
B (0, 1) lies on S=0 ⇒ (0)2+ (1)2+2g (0) +2f (1) +c=0 ⇒ 2f +k =-1………..②
sub f value in eq’’n ④
-2g+4f =3
-2g+4(
-2g-6f=8
-2g=3+2
10f=-5
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)} 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=0 ⇒4g +k=-4………..①
C (4, 5) lies on S=0 ⇒ (4)2+ (5)2+2g (4) +2f (5) +c=0 ⇒8g + 10f +k =-41……….③
) =3
⇒ g=
⇒f=Sub the value of g and f in eq’’n ① ⇒2 ( ) + 2(- ) +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)} 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………..① B (-6, 0) lies on S=0 ⇒ (-6)2+ (0)2+2g (-6) +2f (0) +c=0 ⇒-12g +c =-36………..②
Solving eq’’n ① & ② 4g + 0 + k =-4 0 + 2f +k =-1 4g - 2f =-3…………④ Solving eq’’n ④ & ⑤
4g-2(
-4g-4f=20
) =-3
4g=-3-
-6f=17
⇒ g=
⇒f=-
⇒g=-
Sub the value of g and f in eq’’n ① ⇒4g +k=-4 ⇒4 (- ) +k=-4
x2+y2+2(
Solving eq’’n ④ & ⑤ 14g+2f =34 4g+2f=14 10g=20 ⇒g=2
sub f value in eq’’n ④
4g-2f =-3
∴the required eq’’n of the circle is
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 2g + 2f +c =-2 -12g +c =-36 14g+2f =34…………④
②&③ 0 + 2f + k =-1 8g + 10f +k =-41 -8g-8f=40……….…⑤
)x+2(
)y+ =0.
3x2+3y2-13x-17y+14=0
②&③ -12g +0 +c =-36 -4g + 4f +c =-8 -8g-4f=-28…….…⑤ sub f value in eq’’n ④ 14g+2f =34 ⇒ 14(2) +2f =34 ⇒ 2f= -28 f=3
⇒ ∴ given 4 points are concyclic, D (0, c) lies on the above circle ⇒3(0)2+3(c)2-13(0)-17(c)+14=0 ⇒3c2-17c+14=0 ⇒3c2-3c-14c+14=0 ⇒3c(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 circle passing through the points (4, 1), (6, 5) and whose centre lies on
Aims tutorial 2B important questions & Ans {H/W (2, -3), (-4, 5) and 4x+3y+1=0, (4, 1), (6, 5), and 4x+y16=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=0 ⇒8g + 2f +c =-17………..①
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
Centre C1= (3, 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 =-17 12g +10f+c =-61
C2 (-1, 4)
And r = √ ⇒r1=√( )
( )
√( )
( )
-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 eq’’n ③& ④
)
√(
-4g-8f = 44 4g+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=15 ∴Required eq’’n 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=0 ⇒8g +10f+c =-41………..② Solving eq’’n 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
(
)
=5
√
∴given two circles touch externally The point of contact P divides P=*
(
+=*
=*
+
*
)
( )
( )
( )
+
+
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
) =0
⇒-8x+6y-12=0 Or 4x-3y+6=0.
7)
a) x2+y2-4x-6y-12=0, x2+y2+6x+18y+26=0.(H/W) 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 = √
⇒g=-
⇒r1=√(
Substituting f=0,g=-7/3 in eq’’n (1) -4g + 6f +c =-13
) √(
(
) )
√
√ ( )
√
√
-4( ) +6(0) +c=-13 C=-13C= ∴Required eq’’n 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 find the point of contact and common tangent at this point of contact.
√(
)
(
)
√
There exist two direct common tangents
=√
Aims tutorial 2B important questions & Ans External centre of similitude ‘Q’ divides
⇒m=
, m=
∴the equation of the tangent is (y+4) = ( )(x-22) Q=*
(
+=*
)
(
(
)
)
⇒24(y+4) =7(x-22)
( )
+
⇒24y+96= 7x-154⇒7x-24y-250=0 =*
+
*
+ And is (y+4) = (
The equation to the pair of direct common tangents through Q(22, -4) to the circle S is (
)
(
)
⇒4(y+4) =-3(x-22) ⇒4y+16= -3x+66⇒3x+4y-50=0
Where (
)=(22, -4)
8)
⇒ ⇒ (
)
(
)
(
=
) (
( )
) (
)(x-22)
)
(
)
(
)
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
⇒ =
Centre C1= (2, 5)
⇒
=
⇒
=
⇒
C2 (-2, 3)
And r = √ ⇒r1=√(
)
(
=
)
√
√( )
⇒{121
(
√
)
√
√
} =
⇒21
=0 )
√(
(
)
=√
√
Method II Let the eq’’n of tangent through Q (22, -4) with slope ‘m’ is There exist two transverse common tangents (Y- )=m(x- ) Enternal centre of similitude ‘P’ divides ⇒(y+4) =m(x-22) mx-y-(22m+4)=0………………. (1) (
The perpendicular distance from ⇒ ⇒
)
Q=*
( )
(
+=*
=*
+
*
)
(
)
(
)
( )
+
+
√ (
(
)
√( )
)
(
⇒
)
√
√( )
The equation to the pair of direct common tangents through Q(22, -4) to the circle S is (
|11m+2|=5√ {S.O.B}
⇒(
) =25(
)
(
) Where (
)
⇒121
⇒
⇒96
⇒ (
⇒96
=
⇒24m (4m+3)-7(4m+3) =0
⇒
⇒(24m-7)(4m+3)=0
=
)
(
)
(
) (
( )
)=(22, -4)
) (
)
(
)
(
)
Aims tutorial 2B important questions & Ans ⇒
=
S x2+y2+2gx+2fy+c=0 and the chord of contact of p with
⇒
=
respect to S=0 is
(
)
, where r is the radius of the circle.
13) ⇒
=
⇒{121
} =
⇒21
=0 1 2
Method II
Find centre and radius of circles given by
1 m2
Let the eq’’n of tangent through Q (22, -4) with slope ‘m’ is (Y- )=m(x- ) Sol:
⇒(y+4) =m(x-22)
y 2 2cx 2mcy 0
√ (
The perpendicular distance from
⇒
2
Given equation of the circle
mx-y-(22m+4)=0………………. (1)
⇒
x
)
(
( )
√
√
Centre (-g, -f) = (√
(
)
(
)
√( )
(
⇒
)
)
√
√( )
)
√
And r=√
√
)
√(√
(√
)
|11m+2|=5√ {S.O.B}
⇒(
) =25(
)
=√
=√
(
)
(
)
=c
⇒121 3
⇒96
Find centre and radius of x2+y2+6x+8y-96=0.
⇒96
Sol: Given equation of the circle is x2+y2+6x+8y-96=0.
⇒24m (4m+3)-7(4m+3) =0
Compare with x2+y2+2gx+2fy+c=0.
⇒(24m-7)(4m+3)=0
Centre (-g, -f) = (-3, -4)
⇒m= , m=
Radius r=√
∴the equation of the tangent is (y+4) = ( )(x-22)
=√
⇒24(y+4) =7(x-22)
4
⇒24y+96= 7x-154⇒7x-24y-250=0 And is (y+4) = (
)(x-22)
Find the equation of the circles which touch 2x-3y+1=0 at (1, 1) and having radius √ . 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 are the angles of inclination of tangents through a point p to the circle x2+y2=r2, then find locus of p when cot =k. 12) Show that the area of the triangle formed by the two tangents through p(x1, y1) to the circles
)
(
)
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 √ from (3, 3) ( )
=√
⇒4(y+4) =-3(x-22) ⇒4y+16= -3x+66⇒3x+4y-50=0
=√(
=√
( ) =√
5
Find the power of the point (3, 4) w. r. t the circle x2+y2-4x-6y12=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.
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.
9)
⇒ (
)
Aims tutorial 2B important questions & Ans √ 7
=√
20
( )
√
√ .
If length of tangent from (2,5) to the circle is√ , then find k. Sol: Length of the tangent √ from (2, 5) =√
( )
( ) √
√
∴k=-2 Obtain parametric equation of the circle If x2+y2-6x+4y12=0, x2+y2+6x+8y-96=0. Sol: centre (3, -2), c=-12 And r=√ ( ) =√ Parametric equation of the circle X= ,
9
10
11
12
If x2+y2-4x+6y+c=0 represents a circle with radius 6, find the value of c. Sol: centre (2, -3) and r=√ ( ) =√ S.O.B ⇒13+c=36 ∴=23. If x2+y2-4x+6y+a=0 represents a circle with radius 4, find the value of a. Sol: If x2+y2+6-8y+c=0 represents a circle with radius 6, find the value of c. Sol: 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=-6 Radius r=√
13
14
15
16
17
19
( )
=√
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 ⇒ Find the equation of circle passing through (5, 6) and having centre (-1, 2). Sol: Find the equation of circle passing through (0, 0) and having centre (-1, 2). Sol: Find the equation of the circle passing through (2, 3) and concentric with x
18
√
=√ 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
2
y 2 8 x 12 y 15 0 .
Sol: Find the pole of x + y + 2 = 0 with respect to the circle x2 + y2 – 4x + 6y – 12 = 0. Sol: Show that the points (4, -2), (3, -6) are conjugate w.r.to the circle x2+y2=24. Sol:
2 x 3 y 11 0 and 2 x 2 y 1 0 are conjugate with respect to the circle
Show that the lines
x 2 y 2 4 x 6 y 12 0 .
√
S.O.B
8
21
If (4, k), (2, 3) are conjugate points with respect to the circle x2 + y2 = 17 then find k. Sol:
22
