)
⇒ (-6) + (0) +2g (-6) +2f (0) +c=0 =-36………..(2) (2) ⇒-12g +c =-36……….. C (-2, 2) lies on S=0 ⇒ (-2) + (2) +2g (-2) +2f (2) +c=0 8……….(3) ⇒-4g + 4f +c = - 8……….(3)
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……… +2gx+2fy+c=0……….. (*) (1, 1) lies on S=0 (1)2+ (1)2+2g (1) +2f (1) +c=0 2g + 2f +c =-2………. =- 2……….(1) (1)
⇒ ⇒
(2, -1) lies on S=0 (2)2+ (-1)2+2g (2) +2f (-1) +c=0 4g - 2f +c =-5……….. =-5………..(2) (2)
⇒ ⇒
Solving eq’’n (1) & (2) 2g + 2f +c =-2 4g - 2f +c =-5 -2g+4f =3………… =3…………(4) (4)
(2) & (3) 4g - 2f +c =-5 6g + 4f +c =-13 -2g-6f=8……….… -2g-6f=8……….…(5) (5)
Solving eq’’n (4) & (5)
sub f value in eq’’n (4)
10f=-5
) =3
-2g+4(
-2g=3+2
⇒ g=
⇒f=- Sub the value of g and f in eq’’n e q’’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 H/W (ii)(9,1), (7, 9), (-2, 12), &(6, 10)} Sol: let the equation of of the required circle 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……….. =- 2………..(1) (1)
⇒ ⇒
B (-6, 0) lies on S=0
2
2
Solving eq’’n (4) & (5) 14g+2f =34 4g+2f=14 10g=20 g=2
(2) & (3) -12g +0 +c =-36 -4g + 4f +c =-8 -8g-4f=-28…….… -8g-4f=-28…….… (5) (5) sub f value in eq’’n (4) 14g+2f =34 14(2) +2f =34 2f= -28 f=3
⇒ ⇒
Sub the value of g and f in eq’’n 2 ( ) + 2( ) +c =-2 4+6+c=-2 c=-12
⇒ ⇒
-2g-6f=8
2
Solving eq’’n (1) & (2) 2g + 2f +c =-2 -12g +c =-36 14g+2f =34………… =34………… (4)
⇒
(3, 2) lies on S=0 (3)2+ (2)2+2g (3) +2f (2) +c=0 6g + 4f +c =-13………. =- 13……….(3) (3)
-2g+4f =3
2
⇒ ⇒ ⇒ ∴the required eq’’n of the circle is
(1)
x2+y2+4x+6y-12=0. Now substituting D (-2, -8) in the above eq’’n, eq’’n, we have 2 2 (-2) +(-8) +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 H/w (1, 2), (3, -4), (5, -6), (c, 8)} Sol: let the equation of 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……… +k=-4……….... (1)
⇒ ⇒
B (0, 1) lies on S=0 (0)2+ (1)2+2g (0) +2f (1) +c=0 2f +k =-1……… =-1……….... (2)
⇒ ⇒
C (4, 5) lies on S=0 (4)2+ (5)2+2g (4) +2f (5) +c=0 8g + 10f +k =-41……… =-41……….. (3)
⇒ ⇒
Solving eq’’n (1) & (2) 4g + 0 + k =-4 0 + 2f +k =-1 4g - 2f =-3………… =-3………… (4) (4) Solving eq’’n (4) & (5) 4g-2f =-3 -4g-4f=20 -6f=17
⇒f=-
(2) & (3) 0 + 2f + k =-1 8g + 10f +k =-41 -8g-8f=40……….… -8g-8f=40……….… (5) (5) sub f value in eq’’n (4)
4g-2( ) =-3 4g=-3 ⇒ g= ⇒g=-
Sub the value of g and f in eq’’n e q’’n (1) 4g +k=-4
⇒ ⇒4 (- ) +k=-4 ⇒ ⇒ eq’’n of the circle is ∴the required x +y +2( )x+2( )y+ =0. ⇒ 3x +3y -13x-17y+14=0 ∴ given 4 points are concyclic, D (0, c) lies on the above circle ⇒3(0) +3(c) -13(0)-17(c)+14=0 ⇒3c -17c+14=0 ⇒3c -3c-14c+14=0 ⇒3c(c-1)-14(c-1) =0 ⇒(c-1) (3c-14) =0 ⇒(c-1) =0or (3c-14) =0 ∴c=1, c= 2
2
2
2
2
2
⇒4g+3f+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 x 2+y2-6x-8y+15=0. 5) Find the equation of the circles whose centre 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……… =-13……….... (1)
⇒
⇒ ⇒
⇒ ⇒
⇒ ⇒
∴
⇒ ⇒
2
(4, 5) lies on S=0 (4)2+ (5)2+2g (4) +2f (5) +c=0 8g +10f+c =-41……… =-41……….. .. (2) Solving eq’’n (1) & (2) -4g + 6f +c =-13 8g +10f+c =-41 -12g-4f = 28………. 28……….(3) (3) Given centre (-g, -f) lies on X- axis -f=0 or f=0 -12g-0=28
2
⇒ ⇒
4) Find the equation of the equation of the circle passing 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} 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……….. =- 17………..(1) (1)
⇒ ⇒g=-
Substituting f=0,g=-7/3 f=0,g=-7/3 in eq’’n (1) -4g + 6f +c =-13
⇒ ⇒
(6, 5) lies on S=0 (6)2+ (5)2+2g (6) +2f (5) +c=0 12g +10f+c =-61……… =-61……….... (2) Solving eq’’n (1) & (2) 8g + 2f +c =-17 12g +10f+c =-61
⇒ ⇒
-4g-8f -4g-8f = 44……….(3) 44……….(3) Given centre (-g, -f) lies on 4x+3y-16=0 4 (-g) +3(-f)-24=0
⇒
-4( ) +6(0) +c=-13 C=-13-
C= -
is ∴Requiredeq’’n of the circle x +y +2( )x+2(0)y+( )=0. ⇒3 x +3y -14x-67=0. 2
2
2
2
6) Show that the circles touch each other. Also find the point of contact and common tangent at this point of contact. a) x 2+y 2-6x-2y+1=0, x 2+y 2+2x-8y+13=0. Sol: given equation of the circles
S x2+y2-6x-2y+1=0, and
Centre C1= (3, 1)
C2 (-1, 4)
⇒r = () () () () () () ( ) () √ =5 ∴given two circles touch externally The point of contact P divides 1=0()() ()()1 P=0 1 0 1 =0 And r =
⇒r = () ) () ) √ √ () () () √ √ () ( ) () √ =√ 1
There exist two transverse common tangents Internal centre of similitude ‘P’ divides
1
Equation of common tangent at point of contact is the radical axis of the circles S-S’=0 S-S’=0
) =0
(x2+y2-6x-2y+1=0)- ( x2
⇒-8x+6y-12=0 Or 4x-3y+6=0. 7) Find the equation of the pair of transverse common common tangents to the circles x 2+y 2-4x-10y+28=0, x 2+y 2+4x-6y+4=0. Sol: Sol: given equation of the circles
S x2+y2-4x-10y+28=0, and
Centre C1= (2, 5) And r =
C2 (-2, 3)
)() ()()1 1=0() 0 1 0 10 1 =0 P=
The equation to the pair of transverse common tangents through P(1,9/2) to the circle S is
( ) ( ) Where ( ) = (1,)
⇒ ⇒0( ⇒0() ./ ( ) . /1 =, - 0() () () . ./1 ⇒0 1 =, - 0 1 =, - 01 ⇒0 1 ⇒01 ,-=, - 01 ⇒,-=, ⇒ {4 } =, ⇒3 =0 x 2+y 2-4x-6y-12=0, x 2+y 2+6x+18y+26=0.(H/W)
⇒,-=, -, -, -⇒, ⇒,-=, -, -, -⇒,-=, -,⇒{121 } =, ⇒21 =0
8) Find the equation of direct common tangents to the circles x 2+y 2+22x-4y-100=0, x 2+y 2-22x+4y+100=0. Sol: given equation equation of the circles
S x2+y2+22x-4y-100=0, and
Centre C1= (-11, 2)
C2 (11,-2)
⇒r = () ) () ) √ √ () ) () √ √ () () () √ =√ And r =
Method II
Let the eq’’n of tangent through Q (22, -4) with slope ‘m’ is (Y- )=m(x- )
1
There exist two direct common tangents External centre of similitude ‘Q’ divides
)() () )()1 1=0() 0 1 0 1,=0
Q=
The equation to the pair of direct common tangents through Q(22, -4) to the circle S is
( ) ( ) Where ( )=(22, -4) ⇒ ⇒,() ) () ) ( () ) ( ) - =, -,() () () ()()⇒,- =, -, -
⇒(y+4) =m(x-22) mx-y-(22m+4)=0………………. 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)=0 ⇒m= , m= ∴the equation of the tangent is (y+4) = ( )(x-22) ⇒24(y+4) =7(x-22) ⇒24y+96= 7x-154⇒7x-24y-250=0 And is (y+4) = ( )(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
. √ () () ( ) () √ The centres () ( ( ) () Slope of (1) = - ( ) =2 Slope of line ⇒ √ √ The centres =. √ 0 0√ 1 √ 0 0√ 1/ = {1()} 2x-3y+1=0 at (1, 1) and having radius Sol: given two circles touch the line
= (1-2, 1+3) and (1+2, 1-3) = (-1, 4) and (3, -2)
√ √ ⇒ = √ ⇒ - √ √ S.OB ⇒ ( ) (√ (√ ) ⇒ ( ) ⇒ ) ⇒ ( ) ( ) ( ) Where m , m be the slopes of the tangents which make angles m =tan m =tan =k. Given cot cot ⇒+ ⇒+ ⇒+ ⇒ ⇒ ⇒ ( ) of locus of P(x , y ) is ∴ The equation ( ) Y= This passes through P(x1, y1)
1
1
2
2
1
() () √ () √ ⇒() ⇒ Eq’’n of the circle with ( ( ) ) √ () √ ⇒() ⇒ Eq’’n of the circle with
10) Show that the poles of tangents to the circle x 2+y 2=r 2 w.r.to the circle (x + a) 2+y 2=2a 2 lie on y 2+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=2a2 x 2+y2 +2ax-a2=0 Now the polar of p w.r.t to S=0 is S 1=0 xx1+yy1+a(x+x1) - a2=0 x(x1+a) +yy1+ (ax1- a2) =0……….. (1)
⇒ ⇒ ⇒
: Let P(x1, y1) be the any point on the locus Equation of tangent through p with slope ‘m’ is
() ( ) ⇒( ) *( ) + ⇒ *( ) *( ) + ⇒ ( ) ( ) ⇒ ∴ The equation of locus of P(x , y ) is y +4ax=0 11) If are the angles of inclination of tangents through a point p to the circle x +y =r , then find locus of p when cot =k. 1
2
1
2
2
2
Sol: given equation of the circle x 2+y2=r2………. (1)
1
12) If the chord of contact of a point P with respect respect to
the circle x 2+y 2=a 2 cut the circle at A and B such that AOB =90 0 then show that P lies on the circle x 2+y 2=2a 2. 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=90 0. The equation of the chord of contact of p(x 1, y1) with respect to (1) sis xx1+yy1-a2=0……….. =0……….. (2) The equation to the pair of lines OA and OB is
. / Or ( ) ( ) ⇒( ) ( ) … (3) Since AOB=900, we have the coefficient of () () ∴ ⇒
given by x2+y2-a2
Hence the point p(x1, y1) lies on the circle x2+y2=2a2.
13) Prove that the combined equation equation of the pair of tangents drawn from an external point p ( ) to
x(3)+y(-1)-1(x+3)+2(y-1)=0
⇒ ⇒ ⇒ …. (2) () Required eq’’n of the tangent to (1) and it is parallel to (2) is () ()√ ()√ ⇒ ( ) ()√ () √ () () ⇒ ( ) ()√ () √ √ ⇒ ( ) ()
the circle S=0 is . Sol: given that p ( ) is an external point on S=0
let AB be the chord of contact of P to the circle S=0 and its equation is Let Q ( ) be any point on the locus i.e. , be a point on the pair of tangents the ratio that the line AB divides PQ can be determined in 2ways :
: …….(1) 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)}
(1) PB:QB is equal to (2) The line
14) Show that the area of the triangle formed by the two tangents through p(x1 , y1 ) to the circles S x 2+y 2+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 SAQ: 1. Find the equations of the tangents to the circle x 2+y 2-4x+6y-12=0 which are parallel to Sol: given equation of the circle x 2+y2-4x+6y-12=0 Centre (2, -3) and radius (r)
3. Find the equations of the tangents to the circle x 2+y 2+2x-2y-3=0 which are perpendicular perpendicular to
Sol: given equation of the circle x2+y2+2x-2y-3=0 +2x-2y-3=0 …..(1) Centre (-1, 1) and radius (r)
() () √ () ⇒⇒() Since (3) is tangent to (1) r = distance from (-1, 1) )( () ⇒(() √ ⇒ √ √ ⇒ √ √ ⇒ k = √ Hence required eq’’n of tangents are √ = = The given 0….. (2) Any line to the above line is
4.
() () √ () () ) ⇒d=(( ()())( () √ ∴ r =d, the given straight line touches the given
= = If given line touches the given circle then radius of circle = distance from centre (3, -2) to given line
() () √ ()
= = The given line x+y-8=0….. x+y-8=0….. (1) Any line parallel to the above line is
()
() ) ⇒ ⇒ √ ⇒(() ( √ ⇒ (k-1) = √ √ ⇒k=1√ √ Hence required eq’’n of tangents are x+y+1√ If (2) touches the given circle then r = distance from (2, -3)
2. Find the equation of the tangent to x 2+y 2-2x+4y=0 at (3, -1). Also find the equation of tangent parallel to it. Sol: given equation of the circle circle x2+y2-2x+4y=0 …..(1) Centre (1, -2) and radius (r)
() () √ ()
= = The equation of tangent at (3, -1) is
Show that the line 5x+12y-4=0 touches the circle x 2+y 2-6x+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)
circle. 5.
Show that the tangent at (-1, 2)of the circle x 2+y 2-4x-8y+7=0 touches the circle x 2+y 2+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 circle x circle x +y +4x+6y=0 centre (-2, -3), r = () () () =√ from centre (-2, -3) to given line (1) Distance ( ) ( ( ) ( ) √ ()()()) √ so the line √ (1) also touches the 2 circle. ( ) 2
2
nd
() ) ( ) (() ) () ) ) ⇒ (( ⇒ () ⇒ ⇒
Coordinate of point of contact =(1, -1.) 6. Find the equations of normal to the circle x 2+y 2 4x+6y+11=0 at (3, 2) also find the other point where normal meets the circle Sol: given equation of the circle x2+y2-4x-6y+11=0 -4x-6y+11=0 …..(1) Centre C (2, 3) = (-g, -f) Given point A (3, 2) = ( ) The equation of the normal is
()( )()( ) ⇒()()()() ⇒ ⇒ centre of the circle is mid point of A and B ⇒ 0 1 ( ) ⇒ ⇒ ⇒ ( ) ()
7. Find the area of the triangle formed by the normal normal 2 2 at (3, -4) to the circle x +y +22x-4y+25=0 with the coordinate axes. Sol: given equation of the circle x2+y2-22x-4y+25=0 -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 to x 2+y 2-6x-8y+5=0 sol: given equation of the circle x2+y2-6x-8y+5=0 -6x-8y+5=0 …..(1)
() () √ () The pole =. / () () / =. () )( () ) ( () )( () )
= Centre (3, 4) and r = Given line 3x+4y-45=0 here l=3, m=4 and n=-45
()/ . () () )=(6. 8) =
9. If a point P is moving such that the lengths of tangents from P to the circle x 2+y 2-6x-4y-12=0 and x 2+y 2+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 is 5 x 2+5y 2-78x-108y-212=0
10. Find the length of chord intercepted by the the circle x 2+y 2-8x-2y-8=0 on the line x+y+1=0. Sol: given equation of the circle x2+y2-8x-2y-8=0 -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 =2 11. Find the equation of the circle with centre 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 eq’’n of the circle is () () ⇒() () x2+y2+4x-6y+8=0
12. Find the equation of pair of tangents drawn from (0, 0) to x 2+y 2+10x+10y+40=0 Sol: given equation of the circle x2+y2+10x+10y+40=0 …..(1) P(x1, y1= (0, 0)
() () ( ) ( ) 0 +0 +10(0) +10(0) +40=40 ( ) ( x +y +10x+10y+40)() 25() ( x +y +10x+10y+40)() ⇒* + + * ⇒* ⇒ 2
2
Centre (-g, -f) = (-3, -4)
() () = () =√
Radius r=
3
2
2
2
√ = () () =√ =√
2
4
13. Find the value of k if kx+3y-1=0, 2x+y+5=0 are conjugate lines with respect to the circle x 2+y 2-2x-4y-4=0. Sol: condition is
( ) )( ) ( )( ⇒ () ) (() ) () ) )(( (() ) () ) )
⇒9 (2k+3) = (-k-5) (-9) ⇒2k + 3 = k+ 5 ⇒ k = 5- 2 ∴K=2. 1
Sol:
2
x
2
y
2
2cx
2mcy
5
If length of tangent from (2,5) to t o the circle
7
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
Given equation of the circle
) ( ) (
2
Find centre and radius of 3x 2+3y2-6x+4y-4=0. Sol: Given equation of the circle is 3x2+3y2-6x+4y-4=0.
6
And r=
=
⇒ ⇒. / / ( ) = .
0
( ) ⇒ √ √ ) Centre (-g, -f) = (
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 ( ()( ()
=9+16-12-24-12=-23.
VSAQ Find centre and radius of circles given by 1 m
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)
is√ , then find k. Sol: Length of the tangent √ from (2, 5) = () () √ ⇒ √ √ S.O.B ⇒ ∴k=-2 () √
= And r= Parametric equation of the circle X= ,
= ()=c () Find centre and radius of x 2+y2+6x+8y-96=0.
Sol: Given equation of the circle is x 2+y2+6x+8y-96=0. Compare with x2+y2+2gx+2fy+c=0.
8
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. 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 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=-6 Radius 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) 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) and radius √ is () () (√ ) ) ⇒x +y +2x-4y+5-52=0. 2
2
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), - 3), point on the circle P(0, 0) Radius (r) =CP=
( ) ()=√ ()
√ with centre (-4, -3) and ∴ equation of the circle radius is () () () ⇒x +y +8x+6y+16+9-25=0. ∴x +y +8x+6y=0. 2
2
2
2
16 Find the equation of the circle passing through (2, 3) and concentric with x
2
y
2
8 x 12 y
15
0.
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 eq’’n of the circle is x2+y2+8x+12y-65=0, 17 Find the pole of x + y + 2 = 0 with respect to the circle x2 + y2 – 4x + 6y – 12 = 0. Sol: 18 Show that the points (4, -2), (3, -6) are conjugate w.r.to the circle x2+y2=24. Sol:
⇒ ⇒ ∴
⇒
( ) ( ) ⇒ (4) (3) + (-2) (-6)-24 ⇒12+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 x +y +2gx+2fy+c=0. () ) () ) () ) 2
2
