The given differential equation is(3x 4y 1)dx (4x 5y 1)dy = 0 (i)Comparing Eq (i) with Mdx Ndy = 0, we get M = 3x 4y 1and N = 4x 5y 1Here, ∂M∂y = ∂N∂x = 4Hence, Eq (i) is exact and solution is given by ∫3x 4y 1dx ∫5y 1dy = C⇒ 3x22 4xy x 5y22 y C = 0⇒ 3x2 8xy 2x 5y2 2y 2C = 0⇒ 3x2 2 4 xy 2x 5y2 2y C = 0where, C10 Find the order and degree of the differential equation dx2 6/5 dy dx 6 8 Evaluate Evaluate Evaluate sin4 x cos6 x 3 4 5 to the Circle x' 2gx 2fy c = O Find the angle between the circles x2 — 12x — 6y 41 = O and x 2 y 2 4x6y —59=0 Find the equation of the parabola whose focus is S(I, 7) and vertex is A(I, 2) Show that the curves 2x^23y^2=5 and y^2=x^3 are orthogonal at (1,1) and (1,1) Use parametric mode to draw the curves math Two circles of radius 4 are tangent to the graph of y^2=4x at the point (1,2) Find the equation of these two
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X^2+y^2+2gx+2fy+c=0 differential equation
X^2+y^2+2gx+2fy+c=0 differential equation-General Equation of the Second Degree The equation of the form is a x 2 2 h x y b y 2 2 g x 2 f y c = 0 When a, b and h are not simultaneously zero, is called the general equation of the second degree or the quadratic equation in x and y Homogeneous Equation An equation of the form f ( x, y) = 0 is said to be the homogeneous equation of degree n, where n is a positiveFind slope of the CP
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If the circle x^2y^22gx2fyc=0 bisects the circumference of the circle x^2y^22g^(prime)x2f^(prime)yc^(prime)=0 then prove that 2g^(prime)(gg^(prime))2f^(prime)(ff^(prime))=cc ' Updated On 246\ \Rightarrow b y^2 2fy a x^2 2gx 2C = 0\ The above equation represents a circle \\text{ Therefore, the coffecients of }x^2\text{ and }y^2\text{ must be equal These are respectively y = m x c, ( x − h) 2 ( y − k) 2 = r 2, y = ( a x b) c x d, a x 2 2 h x y b y 2 2 f x 2 g y 1 = 0 The geometric conditions are euclidean motion, curvature, curvature variations The number of geometric conditions equals number of arbitrary geometric constants or differentiations to arrive at an
The given equation isx 2 y 2 2gx2fyc=0 (1)Differentiating once with respect to x, we get2x2yy'2g2fy'=0 (2)Differentiating again, we get22yy''2 (y') 2 2fy''=0or, 1yy'' (y') 2 fy''=0 (3)Differentiating once again, we gety'y''yy'''2y'y''fy'''=0or, yy'''3y'y''fy'''=0 (4)Substituting the value of f from equation (2) in equation (4), we get the required differentialAx^22hxyby^22gx2fy c =0, actually represents the general equation of any second degree curve(the second order conics like circle,parabola,ellipse,hyperbola and rectangular hyperbola) So, by putting h=0 we get ax^2by^22gx2fyc= 0 which does represent a circle with centre at(g,f) and radius=sqrt(g^2f^2c) The general equation of a circle is x 2 y 2gx 2fy c = 0 (i) Since it passes through origin (0, 0), it will satisfy equation (i) ⇒ 2(0)2 (0) 2g(0) 2f(0) c = 0 ⇒ c = 0 ⇒ 2 2x y 2gx 2fy = 0 This is the equation of a circle
Then the equation of parabola is given by (y – β) 2 = 4a (x – α) which is equivalent to x = Ay 2 By C If three points are given we can find A, B and C Similarly, when the axis is parallel to the y – axis, the equation of parabola is y = A'x 2 B'x C' Illustration Find the equation of the parabola whose focus is (3 , 40 votes 1 answerSubtract y^ {2} from both sides Subtract y 2 from both sides 2gx2fyc=0\sin (\frac {\mathrm {d}} {\mathrm {d}x} (y))\cos (\pi x)x^ {2}y^ {2} 2 g x 2 f y c = 0 s i n ( d x d ( y)) c o s ( π x) − x 2 − y 2 Subtract 2gx from both sides Subtract 2 g x from both sides
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If the equation `x^2y^22h x y2gx2fyc=0` represents a circle, then the condition for that circle to pass through three quadrants only but not pass asked in Mathematics by JohnAgrawal (910k points) class12;The parametric equation of the circle x2 y2 2gx 2fy c = 0 is x = g rcosθ, y = f rsinθ Here, θ is a parameter, which represents the angle made by the line, joining the point (x, y) with the center, with the X axis That's it for this lesson See you in the next one! If L = l x m y n = 0 is tangent to circle x 2 y 2 2 g x 2 f y c = 0 then find point of contact I know that equation of tangent is S 1 = x x 1 y y 1 g ( x x 1) f ( y y 1) c = 0 anf rearranging I obtain S 1 = ( g x 1) x ( f y 1) y g x 1 f y 1 c = 0 By comparing coefficients it should give the point of contact
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If The Two Circle X 2 Y 2 2gx C 0 And X 2 Y 2 2fy C Have Equal Radius Askiitians
Key Point The general equation of a circle is x 2y 2gx2fy c = 0, where the centre is given by (−g,−f) and the radius by r = p g2 f2 − c The equation can be recognised because it is given by a quadratic expression in both x and y with no xy term,Answer (1 of 6) The equation x^2y^22gx2fyc=0 comes from the definition of the circle itself The definition of circle is "All the points that are equidistant from a single point" & tge distance is the radius of the circle Consider a circle with radius 'r' & let its center be located at (gThe general equation of conics of a second degree is given by \(a{x^2} 2hxy b{y^2} 2gx 2fy c = 0\) and discriminant Δ = abc 2fgh – af 2 – bg 2 – ch 2 The above given equation represents a nondegenerate conics whose nature is given below in the table
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If P(x, y) be a variable point on the parabola then PS = PM in the figure and this yields the equation to the parabola 2 2 (hx ky)2 hk2 h2 k = x− 2 y− 2 , h2 k2 h k2 h k2 On simplification, the above equation reads y x 2 y x − 1 = 0, −2 k h k h or more compactly, x y ± ± = 1, h and k being parameters h k CarryingLet C_1x^2y^22gx2fyc=0 and C_2=x^2y^22gx2fyc_1=0 Now note that C_1=(xg)^2(yf)^2=f^2g^2c and C_2=(xg)^2(yf)^2=f^2g^2c_1 Now assuming that c\ge c_1, C_1 is contained inSolution For The differential equation which represents the three parameter family of circlesx2y22gx2fyc=0 is The differential equation which represents the three parameter fam Filo About Us Become a Tutor Blog Download App
The Order Of The Differential Equation Whose Solution Is X 2 Y 2 2gx 2fy C 0 Is
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Equation of Tangent to a Circle A tangent to a circle is a straight line which intersects (touches) the circle inexactly one point Let the equation of the circle be x 2 y 2 2gx 2fy c = 0 Let P(x 1, y 1) be a given point on it Let PT be the tangent at P The centre of the circle is C(− g, − f) How To Derive?To ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW If the straight lines `ax^2 2hxy by^2 2gx 2fy c = 0` intersect on the XThe General Form of the equation of a circle is x 2 y 2 2gx 2fy c = 0 The centre of the circle is (g, f) and the radius is \(\rm \sqrt{g^2f^2c}\)
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The GS is y^2 = (A x^4)/(2x^2) Or, alternatively y = sqrt(A x^4)/(sqrt(2)x) We have (x^2 y^2) \\ dx xy \\ dy = 0 Which we can write in standard form as dy/dx = (x^2 y^2)/(xy) 1 Which is a nonseparable First Order Ordinary Differential Equation A suggestive substitution would be to perform a substitution of the form y = xv => dy/dx = v xv' \\ \\ \\ whereFirst order differential equation (dy dx)4 (dy dx)2 − 5x = 0 The maximum derivative of y with respect to x is dy dx Second order differential equation d2y dx2 7y = 0 The maximum derivative of y with respect to x is d2y dx2 Third order differential equation∎ If S ≣ x 2 y 2 2gx 2fy c = 0 and S' ≣ x 2 y 2 2g'x 2f'y c' = 0 are two intersecting circles, then S λS' = 0, λ ≠ −1 , is the equation of the family of circles passing through the points of intersection of S = 0 and S' = 0 ∎ If S ≣ x 2 y 2 2gx 2fy c = 0 is a circle which is intersected by the straight line μ ≣ ax by c = 0 at two
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If the equation `ax^(2)2hxyby^(2)2gx2fyc=0` jointly represents two lines whose separation equations are `xy=0` and `2x3y=1` then `g=` asked 2 days ago in Straight Lines by Somyek (The general equation of a circle is given by x 2 y 2 2gx 2fy c = 0 We will discuss some of the ways of finding the family of circles under the given certain conditions Let S = x 2 y 2 2gx 2fy c = 0 S 1 = x 2 y 2 2g 1 x 2f 1 y c 1 = 0 S 2 = x 2 y 2 2g 2 x 2f 2 y c 2 = 0 L = lx my n = 0 1 Family of2hxy 2gx 2fy c = 0, the origin is to be shifted to the point In this case, the transformed equation is 2hXY c = 0 7 The xy term is removed from ax 2 2hxy by 2 2gx 2fy c = 0 by rotation of axes through an angle θ = 8The condition that the equation ax 2 2hxy by 2 2gx 2fy c = 0,
If The Equation Of The Circle Obtained Byreflecting The Circle X 2 Y 2 A 2 0 In The Liney Mx C Is X 2 Y 2 2gx 2fy C 0 Then A G Frac 2mc 1 M 2 C A 2 B G Frac 2mc 1 M 2 F Frac 2c 1 M 2 C G Frac 2mc 1 M 2 Frac 4c 2
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For the differential equation d x d y 1 x 3 3 x 2 y = 1 x 3 sin 2 x The solution is y ( 1 x 3 ) = A 1 x − B 1 sin 2 x c , where c is the constant of integration, then find B/A?Find eqn to the cone with vertex at origin and base is x2y2 = 4 and z=2 (this is a guiding curve, circle lies on the xy plane) Try this same as example (5) 7 Find eqn to the cone with vertex at (0,0,0) which passes through the curve of intersection of x2 y2z2 x 2y 3z 4 = 0 and xyz=2 Try this also Homogenize these two eqnsWhere g , f & c are arbitary constants
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The general solution of the differential equation of all circles having centre at A( 1, 2) is x 2 y 2 2x 4y c = 0 Explanation General equation of a circle with centre ( g, f) is x 2 y 2 2gx 2fy c = 0 If ( g, f) = A ( 1, 2) Then equation of circle, x 2 y 2 2x 4y c = 0Let the equation of the circle be x 2 y 2 2gx 2fy c = 0 If it passes through (0, 0), then c = 0 The equation of circle is x 2 y 2 2gx 2fy = 0 Since the centre of the circle lies on y–axis then g = 0 The equation of the circle is x 2 y 2 2fy = 0(i) This represents family of circles Dertermine the equation with center (h,k) Example 2 Find the center and the radius of the circle x 2 y 2 5x6y5=0 Comparing with the general equation, x 2 y 2 2gx2fyc=0 g=5/2 f=3 c=5 Hence,the center is (5/2,3) and the radius is 14
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Obtain the differential equation of the family of circles x^2 y^2 2gx 2fy c = 0 ;The Order Of The Differential Equation Whose Solution Is X 2 Y 2 2gx 2fy C 0 Is The order of the differential equation whose solution is x 2 y 2 2gx 2fy c = 0 is 1) 1The line x = y touches a circle at the point (1, 1) If the circle also passes through the point (1, 3), then its radius is
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JEE Main & Advanced Mathematics Differential Equations Question Bank Order and degree of differential equations question_answer The order of the differential equation whose solution is \{{x}^{2}}{{y}^{2}}2gx2fyc=0\, is MP PET 1995Example 4 The order of the differential equation whose solution is x2 y2 2gx 2fy c 0, is (a) 1 (b) 2 (c) 3 (d) 4 Solution (c) To eliminate the arbitrary constants g, f and c, we need 3 more equations, that by differentiating the equation 3 times Hence highest order derivative will be 3 3 dx d yChapter 42 FIRST ORDER DIFFERENTIAL EQUATION 5 Hrs Solution of first order variable separable type differential equation Simple Problems Chapter 43 LINEAR TYPE DIFFERENTIAL EQUATION 4 Hrs x 2 y 2gx 2fy c = 0 –––––– (2) Add g 2, f on both sides
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We know that the equation of circle with centre (g, f) and radius √g 2 f 2c is x 2 y 2 2gx 2fy c =0 So, the equation of concentric circle is x 2 y 2 2gx 2fy c' = 0 Here we observe that both the equations have the same centre, but have different radii and c ≠ c' Similarly, a circle with centre (h, k), and the radius r, will have the equation ( x – h ) 2 ( y – k )The order of the differential equation whose solution is \{{x}^{2}}{{y}^{2}}2gx2fyc=0\, is MP PET 1995A collection of circles is called as a system or family of circles There can be various types of family of circles The general equation of a circle is x2 y2 2gx 2fy c = 0 Since this equation involves three unknowns ie g, f and c so we need at least three conditions to get a unique circle
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The Equation Ax 2 By 2 2hxy 2gx 2fy C 0 Represents A Circle Only If
The equation of any conic can be expressed as ax^2 2hxy by^2 2gx 2fy c = 0 ax2 2hxy by2 2gx2f y c = 0 However, the condition for the equation to represent a circle is a = b a = b and h = 0 h = 0 Then the general equation of the circle becomes x^2 y^2 2gx 2fy c = 0 x2 y2 2gx 2f yc = 0Here is the equation of the concentric circle – x 2 y 2 2gx 2fy c = 0 is x 2 y 2 2gx 2fy k = 0 (Equation differs only by the constant term) 2 Contact of Circles When the outer surfaces of two circles are touching, it is known as contact of circles There may be two cases in contact with circles
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