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Primrose Carr
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1 PAIR OF LINES-SECOND DEGREE GENERAL EQUATION THEOREM If the equation then i) S ax + hxy + by + gx + fy + c represents a pair of straight lines abc + fgh af bg ch and (ii) h ab, g ac, f bc Proof: Let the equation S represent the two lines lx 1 + my 1 + n 1 and lx + my + n. Then ax + hxy + by + gx + fy + c ( lx 1 + my 1 + n1)( lx+ my+ n) Equating the co-efficients of like terms, we get ll 1 a, lm 1 + lm 1 h, m 1 m b, and l 1 n + l n 1 g, m1n+ mn1 f, n1n c (i) Consider the product ( h)( g)( f ) ( l1m+ lm1)( l1n+ ln1)( m1n+ mn1) l l m n + m n + mm l n + l n + nn l m + l m + ll mm nn ( ) ( ) ( ) [( 1 1) 1 1 ] l l [( mn + m n ) mm nn ] + mm [( l n + l n ) ll nn ] + nn lm + l m ll mm + ll mm nn a(4 f bc) + b(4g ac) + c(4h ab) 8fgh 4[ af + bg + ch abc] abc + fgh af bg ch lm 1 + lm1 ii) h ab l l m m ( lm 1 lm1) 4 Similarly we can prove g ac and ( l m l m ) 4 ll mm 4 f bc NOTE : If abc + fgh af bg ch, h ab, g ac and f bc, then the equation S ax + hxy + by + gx + fy + c represents a pair of straight lines

2 CONDITIONS FOR PARALLEL LINES-DISTANCE BETWEEN THEM THEOREM If S ax + hxy + by + gx + fy + c represents a pair of parallel lines then h ab and bg af. Also the distance between the two parallel lines is g ac f bc (or) aa ( + b) ba ( + b) Proof : Let the parallel lines represented by S be lx + my + n 1 -- (1) lx + my + n -- () ax + hxy + gx + fy + c ( lx+ my+ n1)( lx+ my+ n) Equating the like terms l a -- (3) lm h -- (4) m b-- (5) l( n1+ n) g -- (6) m( n1+ n) f -- (7) n1n c -- (8) From (3) and (5), lm Dividing (6) and (7) aband from (4) h l g l g, m f m f a g bg af b f Distance between the parallel lines (1) and () is 1 n ( n1+ n) 4nn 1 n ( l + m ) l + m ab. (4 g / l ) 4c a+ b g ac aa ( + b) (or) or (4 f / m ) 4c f bc ba ( + b) a+ b

3 POINT OF INTERSECTION OF PAIR OF LINES THEOREM The point of intersection of the pair of lines represented by hf bg gh af ax + hxy+ by + gx+ fy+ c when h > ab is, ab h ab h Proof: Let the point of intersection of the given pair of lines be (x 1, y 1 ). Transfer the origin to (x 1, y 1 ) without changing the direction of the axes. Let ( X, Y ) represent the new coordinates of (x, y). Then x X + x1 and y Y + y1. Now the given equation referred to new axes will be a( X x ) h( X x )( Y y ) + by ( + y) + g( X+ x) + f( Y+ y) + c ax + hxy + by + X( ax + hy + g) + Y ( hx1+ by1+ f ) + ( ax + hx y + by + gx + fy + c) Since this equation represents a pair of lines passing through the origin it should be a homogeneous second degree equation in X and Y. Hence the first degree terms and the constant term must be zero. Therefore, ax1+ hy1+ g -- (1) hx1+ by1+ f -- () ax + hx y + by + gx + fy + c -- (3) But (3) can be rearranged as x1( ax1+ hy1+ g) + y1( hx1+ by1+ f) + ( gx1+ fy1+ c) gx1+ fy1+ c --(4) Solving (1) and () for x 1 and y 1 x1 y 1 hf bg gh af ab h hf bg gh af x1 and y 1 ab h ab h Hence the point of intersection of the given pair of lines is hf bg gh af, ab h ab h

4 THEOREM If the pair of lines ax + hxy + by and the pair of lines ax + hxy + by + gx + fy + c form a rhombus then ( a b) fg+ h( f g ). Proof: The pair of lines ax + hxy + by -- (1) is parallel to the lines ax + hxy + by + gx + fy + c -- () Now the equation ax + hxy + by + gx + fy + c +λ ( ax + hxy + by ) Represents a curve passing through the points of intersection of (1) and (). Substituting λ 1, in (3) we obtain gx + fy + c...(4) Equation (4) is a straight line passing through A and B and it is the diagonal AB hf bg gh af The point of intersection of () is C, ab h ab h Slope of gh af OC hf bg Slope of Slope of 1 In a rhombus the diagonals are perpendicular ( OC)( AB) gh af g 1 hf bg f g h afg hf bfg ( a b) fg+ h( f g ) g f fg a b h

5 THEOREM If ax hxy by + + be two sides of a parallelogram and px + qy 1 is one diagonal, then the other diagonal is ybp ( hq) xaq ( hp) proof: Let P(x 1, y 1 ) and Q(x, y ) be the points where the digonal px + qy 1 meets the pair of lines. OR and PQ biset each other at M ( α, β ). x + x 1 1 α and β y + y ax hxy by Eliminating y from (1) and px + qy 1 -- () 1 px 1 px ax + hx + b q q x ( aq hpq + bp ) + x( hp bp) + b The roots of this quadratic equation are x 1 and x where ( hq bp) x1+ x aq hpq bp ( bp hq) α ( aq hpq + bp ) Similarly by eliminating x from (1) and () a quadratic equation in y is obtained and y 1, y are its roots where ( hp aq) y + y aq hpq np 1 ( aq hp) β ( aq hpq + bp )

6 Now the equation to the join of O(, ) and (, ) M α β is ( y )( α ) ( x )( β ) α y β x Substituting the values of α and β, the equation of the diagonal OR is ybp ( hq) xaq ( hp). I EXERCISE 1. Find the angle between the lines represented by Sol. Given equation is x + xy 6y + 7y Comparing with x + xy 6y + 7y. ax + hxy + by + gx + fy + c then a, b - 6, c -, g, f 7 1, h Angle between the lines is given by cos α a+ b ( ) a b + 4h ( ) α cos Prove that the equation x + 3xy y +3x+y+1 represents a pair of perpendicular lines. Sol. From given equation a, b - and a + b + (-) angle between the lines is 9. The given lines are perpendicular.

7 II 1. Prove that the equation 3x + 7xy + y + 5x + 5y+ represents a pair of straight lines and find the co-ordinates of the point of intersection. Sol. The given equation is 3x +7xy+y + 5x + 5y + Comparing with ax + hxy + by + gx + fy + c, we get a 3, b, c, f 3 5 f g 5 5 g, h 7 7 h abc + fgh af bg ch ( )( ) ( ) 1 ( 3 3 ) h ab 3. 6 > f bc. 4 > g ac 3. 6 > 4 4 The given equation represents a pair of lines. hf bg gh af The point of intersection of the lines is, ab h ab h

8 ,, ,, Point of intersection is 3 1 p, 5 5. Find the value of k, if the equation x + kxy 6y + 3x + y + 1 represents a pair of straight lines. Find the point of intersection of the lines and the angle between the straight lines for this value of k. Sol. The given equation is x + kxy 6y + 3x + y + 1 a, b - 6, c 1, f 1 3 k,g 3 g, h Since the given equation is representing a pair of straight lines, therefore abc + fgh af bg ch 1 3 k 1 9 k k + 54 k k + 3k+ 4 k 3k 4 (k 4) (k + 1) k 4 or 1 Case (i) k - 1 hf bg gh af Point of intersection is, ab h ab h

9 ,, ,, Point of intersection is 5 1, 7 7 Angle between the lines cos α a+ b 6 ( ) a b + 4h ( ) Case (ii) k ,, Point of intersection is 5 1 P, and angle between the lines is 8 8 cos α a+ b ( ) a b + 4h ( ) α cos 5 1 1

10 3. Show that the equation x y x + 3y - represents a pair of perpendicular lines and find their equations. Sol. Given equation is x y x + 3y - a 1, b 1, c - f 3, 1 g, h Now abc + fgh af bg ch 9 1 1( 1)( ) ( ) h ab 1 1 1> 9 1 f bc > g ac + > 4 4 And a +b 1 1 The given equation represent a pair of perpendicular lines. Let + ( x+ y+ c )( x y+ c ) x y x 3y 1 Equating the coefficients of x c1+ c 1 Equating the co-efficient of y c1+ c 3 Adding c1 c 1 c1+ c 1 c1+ 1 1, c1 Equations of the lines are x + y and x y + 1

11 4. Show that the lines concurrent. x + xy 35y 4x + 44y 1 are 5x + y 8 are Sol. Equation of the given lines are x + xy 35y 4x + 44y 1 a 1, b - 35, c - 1, f, g -, h 1 hf bg gh af Point of intersection is, ab h ab h ,,, Point of intersection of the given lines is 4 P,. Given line is 5x y Substituting P in above line, x + y P lies on the third line 5x + y 8 The given lines are concurrent. 5. Find the distances between the following pairs of parallels straight lines : i). 9x 6xy + y + 18x 6y + 8 Sol. Given equation is 9x 6xy + y + 18x 6y + 8. From above equation a 9,b1,c8,h -3,g9,f-3. g ac Distance between parallel lines a a + b ( )

12 ( + ) ii. ans. x + 3xy+ 3y 3x 3 3y Show that the pairs of lines 3x + 8xy 3y and form a squares. Sol. Equation of the first pair of lines is 3x + 8xy 3y ( x+ 3y)( 3x y) 3x y, x + 3y 3x + 8xy 3y + x 4y 1 Equations of the lines are 3x y..(1)and x + 3y..() Equation of the second pair of lines is 3x + 8xy 3y + x 4y + 1 Since 3x + 8xy 3y ( x + 3y)( 3x y) Let ( 3x y+ c )( x+ 3y+ c ) 3x 8xy 3y x 4y 1 1 Equating the co-efficient of x, we get c 1 + 3c Equation the co-efficient of y, we get 3c 1 + c - 4 c1 c c1 1,c Equations of the lines represented by 3x + 8xy 3y + x 4y + 1 are

13 3x y 1.(3)and x + 3y + 1..(4) From above equations, lines (1) and (3) are parallel and lines () and(4) are parallel. Therefore given lines form a parallelogram. But the adjacent sides are perpendicular, it is a rectangle.( since,(1),() are perpendicular and (3),(4) and perpendicular.) The point of intersection of the pair of lines 3x + 8xy 3y is O(,). Length of the perpendicular from O to (3) Length of the perpendicular from O to (4) Therefore, O is equidistant from lines (3),(4). Therefore, the distance between the parallel lines is same. Hence the rectangle is a square. III 1. Find the product of the length of the perpendiculars drawn from (,1) upon the lines 1x + 5xy + 1y + 1x + 11y + Sol. Given pair of lines is 1x + 5xy + 1y + 1x + 11y + Now 1x + 5xy + 1y 1x + 16xy + 9xy + 1y 4x ( 3x + 4y) + 3y( 3x + 4y) ( 3x + 4y)( 4x + 3y) Let ( 3x + 4y + c )( 4x + 3y + c ) 1x 5xy 1y 1x 11y 1 Equating the co-efficient of x, y we get 4c1+ 3c 1 4c1+ 3c 1.(1)

14 3c1+ 4c 11 3c1+ 4c 11.() Solving, c1 c c1 1,c 7 7 Therefore given lines are 3x + 4y (3) and 4x + 3y (4) Length of the perpendicular form P(,1) on () Length of the perpendicular from P(,1) on ( ) Product of the length of the perpendicular. Show that the straight lines y 4y+ 3 and from a parallelogram and find the lengths of its sides. x + 4xy+ 4y + 5x+ 1y+ 4 Sol. Equation of the first pair of lines is y 4y 3 y 1 y 3 +, ( )( )

15 y 1 ory 3 Equations of the lines are y 1..(1) and y 3..() Equations of (1) and () are parallel. Equation of the second pair of lines is ( ) ( ) x+ y + 5 x+ y + 4 ( ) ( ) ( ) x+ y + 4 x+ y + x+ y + 4 ( x+ y)( x+ y+ 4) + 1( x+ y+ 4) ( x+ y+ 1)( x+ y+ 4) x + 4xy+ 4y + 5x+ 1y+ 4 x+ y+ 1,x+ y+ 4 Equations of the lines are x+ y+ 1..(3)and x+ y+ 4 (4) Equations of (3) and (4) are parallel. Solving (1), (3) x + + 1, x - 3 Co-ordinates of A are (-3, 1) Solving (), (3) x , x - 7 Co-ordinates of D are (-7,3)

16 Solving (1), (4) x + + 4, x - 6 Co-ordinates of B are (-6, 1) ( ) ( ) AB ( ) ( ) AD Length of the sides of the parallelogram are 3, 5 3. Show that the product of the perpendicular distances from the origin to the pair of straight lines represented by c ( ) a b + 4h ax + hxy + by + gx + fy + c is Sol. Let l 1 x+ m 1 y+ n 1.(1) l x+ m y+ n..()be the lines represented by ax + hxy + by + gx + fy + c ax + hxy + by + gx + fy + c ( lx+ m y+ n )( l x+ m y+ n ) ll a,m m b, lm + l m h ln + l n g,m n + m n f,n n c Perpendicular from origin to (1) l n 1 + m 1 1 Perpendicular from origin to () l n + m Product of perpendiculars

17 n1 n. l + m l + m 1 1 nn 1 ll + m m + lm + lm nn 1 ( ll mm ) + ( lm + l m) c c a b + h a b + 4h ( ) ( ) ( ) 4. If the equation ax + hxy + by + gx + fy + c represents a pair of intersecting lines, then show that the square of the distance of their point of intersection from the origin is ( + ) c a b f g ab h. Also show that the square of this distance from origin is f h + g + b if the given lines are perpendicular. Sol. Let l 1 x+ m 1 y+ n 1.(1) l x+ m y+ n..() be the lines represented by ax + hxy + by + gx + fy + c ax + hxy + by + gx + fy + c ( lx m y n )( l x m y n ) ll a,m m b, lm + l m h ln + l n g,m n + m n f,n n c Solving (1) and () x y 1 mn mn l n ln lm l m

18 mn 1 mn 1 ln1 l1n The point of intersection is P, l1m lm1 l1m lm1 OP ( mn 1 mn 1) + ( ln1 l1n) ( lm l m ) 1 1 ( ) ( ) ( lm + l m ) 4ll m m l1 + l 1 ll 1 1 mn mn 4mmnn n n 4 nn h 4ab 4f 4abc 4g 4ac ( + ) ca b f g ab h. If the given pair of lines are perpendicular, then a + b a b OP f g f + g b b h h + b ( ) HOMOGENISATION THEOREM The equation to the pair of lines joining the origin to the points of intersection of the curve S ax + hxy + by + gx + fy + c and the line L lx+ my+ n is lx + my lx + my ax + hxy + by + (gx + fy) + c n n ---(1) Eq (1) represents the combined equation of the pair of lines OA and OB.

19 EXERCISE I 1. Find the equation of the lines joining the origin to the point of intersection of x + y 1 and x+ y 1 Sol. The given curves are x + y 1..(1) x+ y 1..() Homogenising (1) with the help of () then x + y i.e. xy xy ( ) x y x y x y xy. Find the angle between the lines joining the origin to points of intersection of y x and x+ y 1. Sol. Equation of the curve is y Harmogonsing (1) with the help of () Y x.1 ( ) y x x+ y x + xy x..(1) and Equation of line is x+ y 1.() II x + xy y which represents a pair of lines. From this equation a+ b 1 1 The angle between the lines is Show that the lines joining the origin to the points of intersection of the curve x xy+ y + 3x+ 3y and the straight line x y are mutually perpendicular.

20 Sol. Le t A,B the the points of intersection of the line and the curve. Equation of the curve is x xy+ y + 3x+ 3y.(1) Equation of the line AB is x y x y x y 1.() Homogenising, (1) with the help of () combined equation of OA, OB is x xy + y + 3x.1+ 3y.1.1 ( ) ( ) x y x y x xy+ y + 3 x+ y 3 x xy+ y + x y x xy+ y ( ) ( ) 3 3 x xy+ y + x y x + xy y x xy y 3 3 a+ b OA, OB are perpendicular.

21 . Find the values of k, if the lines joining the origin to the points of intersection of the curve perpendicular. x xy + 3y + x y 1 and the line x + y k are mutually Sol. Given equation of the curve is S x xy+ 3y + x y 1 (1) Equation of AB is x + y k x+ y 1..() k Le t A,B the the points of intersection of the line and the curve. Homogenising, (1) with the help of (), the combined equation of OA,OB is x xy + 3y + x.1 y.1 1 ( + ) ( + ) x y x y x xy + 3y + x y k k ( x+ y) k k x k xy + 3k y + kx ( x + y) ( ) ( ) ky x + y x + y k x k xy 3k y kx kxy kxy ky x 4xy 4y ( k + k 1) x + ( k + 3k 4) xy ( ) + 3k k 4 y Given that above lines are perpendicular, Co-efficient x + co-efficient of y k + k 1+ 3k k 4 5k 5 k 1 k ± 1

22 3. Find the angle between the lines joining the origin to the points of intersection Sol. of the curve x + xy+ y + x+ y 5 and the line 3x y + 1 Equation of the curve is x + xy+ y + x+ y 5 (1) Equation of AB is 3x y + 1 y 3x 1 () Le t A,B the the points of intersection of the line and the curve. Homogenising (1) with the help of (), combined equation of OA, OB is x + xy + y + x.1+ y x + xy + y + x ( y 3x) ( ) ( ) x xy y xy 6x y 6xy + y y 3x 5 y 3x 5 y + 9x 6xy ( ) 5x xy + 3y 5y 45x + 3xy 5x + 8xy y 5x 14xy + y let θ be the angle between OA and OB,then a+ b 5+ 1 cos θ a b + 4h ( ) ( ) θ cos 193

23 III 1. Find the condition for the chord lx+ my 1of the circle Sol. centre is the origin) to subtend a right angle at the origin. x + y a (whose Equation of the circle x + y a.(1) Equation of AB is l x+ my 1 () Let A,B the the points of intersection of the line and the curve Homogenising (1) with the help of (),the combined equation of OA, OB is + x + y a ( l x+ my) x y a.1 ( l l ) + + a x m y mxy l + + l a x a m y a mxy a l x + a l mxy+ a m y x y ( l ) l ( ) a 1 x + a mxy+ a m 1 y Since OA, OB are perpendicular, Coefficient of x + co-efficient of y a 1+ a m 1 l ( ) a l + m which is the required condition

24 . Find the condition for the lines joining the origin to the points of intersection of Sol. the circle x + y a and the line lx+ my 1to coincide. Equation of the circle is x + y a (1) Equation of AB is l x+ my 1.(). Le t A,B the the points of intersection of the line and the curve. Homogenising (1) with the help of (), Then the combined equation of OA, OB is x + y a.1 ( ) x + y a x+ my l a ( l x + m y + l mxy) x + y a l x + a m y + a l mxy ( l ) l ( ) a 1 x + a mxy+ a m 1 y Since OA, OB are coincide h ab l l a 4 l m a 4 l m a l a m + 1 a 4 m ( a 1)( a m 1) + a a m 1 l ( l ) a + m 1 This is the required condition.

25 3. Write down the equation of the pair of straight lines joining the origin to the points of intersection of the lines 6x y + 8 with the pair of straight lines 3x + 4xy 4y 11x + y + 6. Show that the lines so obtained make equal angles with the coordinate axes. Sol. Given pair of line is 3x + 4xy 4y 11x + y + 6 (1) Given line is 6x y 6x y y 6x () 8 Homogenising (1) w.r.t () y 6x y 6x 3x + 4xy 4y ( 11x y) y + 36x 1xy 64 3x 4xy 4y 8 11xy 66x y 1xy 936x + 56xy 56xy 34y 468x 117y 4x y ---- (3) Is eq. of pair of lines joining the origin to the point of intersection of (1) and (). The eq. pair of angle bisectors of (3) is ( ) ( ) x y 4 1 xy xy ( ) ( ) h x y a b xy x or y which are the eqs. is of co-ordinates axes The pair of lines are equally inclined to the co-ordinate axes

26 4. If the straight lines given by Y-axis, show that fgh bg ch ax + hxy + by + gx + fy + c intersect on Sol. Given pair of lines is ax + hxy + by + gx + fy + c Equation of Y-axis is x then equation becomes by + fy + c (1) Since the given pair of lines intersect on Y axis, the roots or equation (1) are equal. Discriminate ( ) f 4.b.c 4f 4bc f bc f bc Since the given equation represents a pair of lines abc + fgh + af bg ch ( ) + a f fgh af bg ch fgh bg ch 5. Prove that the lines represented by the equations x+ y 3 form an equilateral triangle. x 4xy+ y and Sol. Since the straight line L: x+ y 3makes 45 with the negative direction of the X axis, none of the lines which makes6 with the line L is vertical. If m is the slope of one such straight line, then m tan6 1 m and so, satisfies the equation( m+ 1) 3( m 1) Or m 4m+ 1...(1)

27 I B 35 3 A I 1 x+ y 3 But the straight line having slope m and passing through the origin is y mx... () So the equation of the pair of lines passing through the origin and inclined at 6 with the line L is obtained by eliminating m from the equations (1) and (). Therefore the combined equation of this pair of lines is y y (i.e,) x x x 4xy+ y Which is the same as the given pair of lines. Hence, the given traid of lines form an equilateral triangle. 6. Show that the product of the perpendicular distances from a point ( α, β ) to the pair of straight lines ax + hxy + by is Sol. Let ax + hxy+ by ( l x+ m y)( l x+ m y) 1 1 aα + hαβ + bβ ( ) a b + 4h Then the separate equations of the lines represented by the equation ax + hxy + by are L1: l1x+ m1y and L : lx+ my Also, we have ll 1 a; mm 1 b and lm 1 + lm 1 h d length of the perpendicular from (, ) 1 α β to L 1 lα + mβ m1 l d length of the perpendicular from (, ) α β to L L l α + m β + m l

28 Then, the product of the lengths of the perpendiculars from ( α, β ) to the given pair of lines dd 1 ( l )( ) 1α + m1β lα + mβ ( l1 + m1 )( l + m) aα + hαβ + bβ ( ) a b + 4h PROBLEMS FOR PRACTICE. 1. If the lines xy+x+y+1 and x +ay- 3 are concurrent, find a.. The equation ax + 3xy y 5x+ 5y+ c represents two straight lines perpendicular to each other. Find a and c. 3. Find λ so that x + 5xy + 4y + 3x + y +λ may represent a pair of straight lines. Find also the angle between them for this value of λ. 4. If ax + hxy + by + gx + fy + c represents the straight lines equidistant from the origin, show that f 4 g 4 c( bf ag ) 5. Find the centroid of the triangle formed by the lines 1x xy+ 7 y and x 3y+ 4 ANS. 8, Let ax + hxy + by represent a pair of straight lines. Then show that the equation of the pair of straight lines. i)passing through (, ) x y and parallel to the given pair of lines is o o ( ) + ( )( ) + ( ) ii) Passing through (, ) a x x h x x y y b y y o o o o perpendicular to the given pair of lines is b( x x ) h( x x )( y y ) + a( y y ) o o o o x y and o o

29 7. Find the angle between the straight lines represented by x + 3xy y 5x+ 5y 3 8. Find the equation of the pair of lines passing through the origin and perpendicular to the pair of lines ax + hxy + by + gx + fy + c 9. If x + xy+ y + 4x y+ k represents a pair of straight lines find k. 1. Prove that equation x + xy 6y + 7y represents a pair of straight line. 11. Prove that the equation x + 3xy y x+ 3y 1 represents a pair of perpendicular straight lines. 1. Show that the equation x 13xy 7y + x+ 3y 6 represents a pair of straight lines. Also find the angle between the co-ordinates of the point of intersection of the lines. 13. Find that value of λ for which the equation λx 1xy+ 1y + 5x 16y 3 represents a pair of straight lines. 14. Show that the pair of straight lines 6x 5xy 6y and 6x 5xy 6y + x+ 5y 1 form a square. 15. Show that the equation 8x 4xy+ 18y 6x+ 9y 5 represents pair of parallel straight lines are find the distance between them.

Q1. If (1, 2) lies on the circle. x 2 + y 2 + 2gx + 2fy + c = 0. which is concentric with the circle x 2 + y 2 +4x + 2y 5 = 0 then c =

Q1. If (1, 2) lies on the circle x 2 + y 2 + 2gx + 2fy + c = 0 which is concentric with the circle x 2 + y 2 +4x + 2y 5 = 0 then c = a) 11 b) -13 c) 24 d) 100 Solution: Any circle concentric with x 2 +