Circles JEE 2024 Past Papers- PDF

 INDEX ❖ Standard Equations of Circles ❖ Intercepts made by a circle ❖ Some standard Notations ❖ Position of a Point w.r.t a circle ❖ Family of Circles ❖ Chords of a circle Standard Equations of a Circle  Standard Equations of a Circle Central form of the Eq...

 INDEX ❖ Standard Equations of Circles ❖ Intercepts made by a circle ❖ Some standard Notations ❖ Position of a Point w.r.t a circle ❖ Family of Circles ❖ Chords of a circle Standard Equations of a Circle  Standard Equations of a Circle Central form of the Equation of a Circle Here, centre is (x1, y1) and radius is ‘r’.  Standard Equations of a Circle Central form of the Equation of a Circle Here, centre is (x1, y1) and radius is ‘r’. NOTE Circle with centre at (0, 0) and radius r is x 2 + y 2 = r 2  Standard Equations of a Circle Central form of the Equation of a Circle For example, (5, 3) (a) (1, 0) (b) 2 (1, 2)  Standard Equations of a Circle Central form of the Equation of a Circle For example, (5, 3) centre is (1, 0) and (a) So, the circle’s equation is (1, 0) (x − 1)2 + (y − 0)2 = 25, that is x2 + y2 − 2x − 24 = 0 centre is (1, 2) and circumference = 2𝜋 (b) ⇒ 2 𝜋r = 2𝜋 ⇒ r = 1 (1, 2) 2 So, circle’s equation is (x − 1)2 + (y − 2)2 = 1, that is x2 + y2 − 2x − 4y + 4 = 0  Standard Equations of a Circle NOTE The diameter or the normal of a circle passes through its centre. One of the equation of a circle of radius 13 whose centre lies on x-axis and passes through the point (4, 5), is A x2 + y2 - 32x + 37 = 0 B x2 + y2 + 16x + 87 = 0 C x2 + y2 - 32x - 105 = 0 D x2 + y2 + 16x - 105 = 0 One of the equation of a circle of radius 13 whose centre lies on x-axis and passes through the point (4, 5), is A x2 + y2 - 32x + 37 = 0 B x2 + y2 + 16x + 87 = 0 C x2 + y2 - 32x - 105 = 0 D x2 + y2 + 16x - 105 = 0 Solution: Slide 12 1 abhi rehne do, but future ke liye, in this solution, r,0 ki jagah a,0 krdo Arvind Kalia, 14-12-2023 A circle has radius 3 units and its centre lies on the line y = x - 1. Find the equation of the circle if it passes through (7, 3). Solution: C (t, t - 1) P(7, 3) y=x-1  Standard Equations of a Circle General form of the Equation of a circle Recall A general two degree equation in two variables is represented as  Standard Equations of a Circle General form of the Equation of a circle x2 + y2 + 2gx + 2fy + c = 0 NOTE Condition for a general second degree equation in two variables to represent a circle is a = b and h = 0. If a = b ≠ 1, then we divide the equation by a constant to make both coefficients equal to 1.  Standard Equations of a Circle General form of the Equation of a circle x2 + y2 + 2gx + 2fy + c = 0 where, Centre is (-g , -f) and For example: Centre Radius x2 + y2 - 4x + 4y - 28 = 0  Standard Equations of a Circle General form of the Equation of a circle x2 + y2 + 2gx + 2fy + c = 0 where, Centre is (-g , -f) and For example: Centre Radius x2 + y2 - 4x + 4y - 28 = 0 (2, -2) 6 (-3/4, -1) 1 Equation of the circle concentric with the circle x2 + y2 - 6x + 12y + 15 = 0 and of double its area is A x2 + y2 - 3x + 12y -15 = 0 B x2 + y2 - 3x + 12y -30 = 0 C x2 + y2 - 6x + 12y - 15 = 0 D x2 + y2 - 6x + 12y -20 = 0 Equation of the circle concentric with the circle x2 + y2 - 6x + 12y + 15 = 0 and of double its area is A x2 + y2 - 3x + 12y -15 = 0 B x2 + y2 - 3x + 12y -30 = 0 C x2 + y2 - 6x + 12y - 15 = 0 D x2 + y2 - 6x + 12y -20 = 0 Solution: and centre is (3, -6)  JEE Main 8th April, 2023 Consider a circle C1 : x2 + y2 - 4x - 2y = 𝞪 - 5. Let its mirror image in the line y = 2x + 1 be another circle C2 : 5x2 + 5y2 - 10fx - 10gy + 36 = 0. Let r be the radius of C2. Then 𝞪 + r is equal to _____.  JEE Main 8th April, 2023 Consider a circle C1 : x2 + y2 - 4x - 2y = 𝞪 - 5. Let its mirror image in the line y = 2x + 1 be another circle C2 : 5x2 + 5y2 - 10fx - 10gy + 36 = 0. Let r be the radius of C2. Then 𝞪 + r is equal to _____. Ans: 2 Solution:  Standard Equations of a Circle Some Special Circles Now let us see some special cases. These are generally used to give information indirectly, in the questions.  Standard Equations of a Circle Some Special Circles 1. Circle touching X - axis 2. Circle touching Y - axis Y (0, b) X (a, 0)  Standard Equations of a Circle Some Special Circles 3. Circle touching X - axis at origin 4. Circle touching Y - axis at origin Y Y X O X O  Standard Equations of a Circle Some Special Circles 5. Circle touching both the axes Y Y O X O X Y Y X O X O Radius of the circle touching both the axes and passing through the point (1, 1) is _____. A B C D None of these Radius of the circle touching both the axes and passing through the point (1, 1) is _____. A B C D None of these Solution: Y (1, 1) (a, a) X O  IIT JEE - 2011 The circle passing through the point (-1, 0) and touching the y-axis at (0, 2) also passes through the point A B C D (-4, 0)  IIT JEE - 2011 The circle passing through the point (-1, 0) and touching the y-axis at (0, 2) also passes through the point A B C D (-4, 0) Solution: (h, 2) (0. 2) 2 (- 1, 0) Find the equation of the circle inscribed in a triangle formed by the coordinate axes and the straight line 3x + 4y = 24. Solution: Y B 10 6 O (2, 2) X D A 3x + 4y = 24 8  Standard Equations of a Circle Diametric form of the Equation of a Circle The circle whose diameter endpoints are A(x1, y1) and B(x2, y2) has equation (x − x1) (x − x2) + (y − y1) (y − y2) = 0  Standard Equations of a Circle Diametric form of the Equation of a Circle The circle whose diameter endpoints are A(x1, y1) and B(x2, y2) has equation (x − x1) (x − x2) + (y − y1) (y − y2) = 0 NOTE Basically it’s the sum of two quadratics, one in x, whose roots are the abscissae and one in y, whose roots are the ordinates of the diametric endpoints.  AIEEE 2011 Find the equation of the circle which passes through (1, 0) and (0, 1) and has its radius as small as possible. Solution: The radius will be minimum, if the given points are the end-points of a diameter. Then, equation of circle is (x - 1) (x - 0) + (y - 0) (y - 1) = 0 ⇒ x2 + y2 - x - y = 0  Standard Equations of a Circle Parametric form of the Equation of a Circle Y r θ X O  Standard Equations of a Circle Parametric form of the Equation of a Circle Y (a) x2 + y2 = r2 ⇒ x = r cos θ, y = r sin θ P(θ) r θ X O In particular, a general point on x2 + y2 = 1 is of the form (cosθ, sinθ) for some θ.  Standard Equations of a Circle Parametric form of the Equation of a Circle (b) (x − x1)2 + (y − y1)2 = r2 ⇒ x = x1 + r cosθ, y = y1 + r sin θ If A(cos ⍺, sin ⍺), B(cos β, sin β), C(cos γ, sin γ) are the vertices of a ΔABC, then find the coordinates of its orthocentre Solution: Solution: Intercepts made by a Circle  Intercepts made by a circle y = mx + c A AB is the intercept made by circle on the line y = mx + c.  Intercepts made by a circle y = mx + c A AB is the intercept made by circle on the line y = mx + c. NOTE Whenever a circle makes an intercept on line, always refer to following figure. r Find length of intercept made by circle x2 + y2 − 2x + 4y − 20 = 0 on line 4x − 3y − 10 = 0. Solution:  JEE Main 12th April, 2019 A circle touching x-axis at (3, 0) and making an intercept of length 8 on the y-axis passes through the point A (1, 5) B (2, 3) C (3, 5) D (3, 10)  JEE Main 12th April, 2019 A circle touching x-axis at (3, 0) and making an intercept of length 8 on the y-axis passes through the point A (1, 5) B (2, 3) C (3, 5) D (3, 10) Solution: From the figure, the equation of the circle is (x - 3)2 + (y - 5)2 = 52 3 (3, 5) 8 So, the point (3, 10) will satisfy the equation 4 5 5 (3, 0)  JEE Main 28th June, 2022 If one of the diameters of the circle is a chord of the circle then the value of r2 is equal to  JEE Main 28th June, 2022 If one of the diameters of the circle is a chord of the circle then the value of r2 is equal to Ans: 10 Solution: P r1 Q C r O 2 rods whose lengths are 2a, 2b slide along axes (one on each) in such a way that their extremities are always concyclic. Find the equation of locus of centre of circle. Solution: Y D r 2b F O (h, k) r C X A B E 2a  Intercepts made by a circle Remark 1. Intercept made by x2 + y2 + 2gx + 2fy + c = 0 on the X - axis A B X  Intercepts made by a circle Remark 1. Intercept made by x2 + y2 + 2gx + 2fy + c = 0 on the X - axis A B X (a) g2 − c > 0 ⇒ Circle cuts the X - axis at two distinct points (b) g2 − c = 0 ⇒ Circle touches the X - axis (c) g2 − c < 0 ⇒ Circle does not meet the X - axis  Intercepts made by a circle Remark 2. Intercept made by x2 + y2 + 2gx + 2fy + c = 0 on the Y - axis Y B A  Intercepts made by a circle Remark 2. Intercept made by x2 + y2 + 2gx + 2fy + c = 0 on the Y - axis Y B A (a) f 2 − c > 0 ⇒ Circle cuts the Y - axis at two distinct points (b) f 2 − c = 0 ⇒ Circle touches the Y - axis (c) f 2 − c < 0 ⇒ Circle does not meet the Y - axis  JEE Main 27th July, 2022 If the circle x2 + y2 - 2gx + 6y - 19c = 0, g, c ∈ R passes through the point (6, 1) and its centre lies on the line x - 2cy = 8, then the length of intercept made by the circle on x-axis is A √11 B 4 C 3 D 2√23  JEE Main 27th July, 2022 If the circle x2 + y2 - 2gx + 6y - 19c = 0, g, c ∈ R passes through the point (6, 1) and its centre lies on the line x - 2cy = 8, then the length of intercept made by the circle on x-axis is A √11 B 4 C 3 D 2√23 Solution: Some standard Notations  Some Standard Notations Here, we will be learning some standard notations for general second degree equations in x and y. These notations will be very helpful in upcoming formulae. Primarily there are are three notations S, S1 and T. Let’s see what do they denote.  Some Standard Notations Notations: Any second degree equation in two variables, that is, ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 will be represented as S = 0. As of now, that we are doing circles, so we have S ≡ x2 + y2 + 2gx + 2fy + c  Some Standard Notations 1. S ≡ x2 + y2 + 2gx + 2fy + c 2. Consider a point (x1, y1). Value of S at (x1, y1) is represented by S1 i.e., S1 = x12 + y12 + 2gx1 + 2fy1 + c 3. Consider a point (x1, y1). If in S we replace then we get T, Write S1 and T for the following: (a) S ≡ x2 + 2y2 − 3x + 4y + 3 at the point (1, 2) (b) S ≡ x2 + 2xy +3y at the point (1, 3) Write S1 and T for the following: (a) S ≡ x2 + 2y2 − 3x + 4y + 3 at the point (1, 2) Solution: Write S1 and T for the following: (b) S ≡ x2 + 2xy +3y at the point (1, 3) Solution: Position of a point w.r.t. a Circle  Position of a point w.r.t. a Circle Method 1 Find distance of point P from centre of circle O. OP < r ⇒ P lies inside the circle OP = r ⇒ P lies on the circle OP > r ⇒ P lies outside the circle  Position of a point w.r.t. a Circle Method 1 Find distance of point P from centre of circle O. OP < r ⇒ P lies inside the circle OP = r ⇒ P lies on the circle OP > r ⇒ P lies outside the circle Method 2 S1 < 0 ⇒ P lies inside the circle S1 = 0 ⇒ P lies on the circle S1 > 0 ⇒ P lies outside the circle  Position of a point w.r.t. a Circle Remark Greatest and least distance of a point from a circle. P |OP - r| = least distance of point P from the circle |OP + r| = greatest distance of point P from the circle Find the greatest distance of the point P(10, 7) from the circle x2 + y2 - 4x - 2y - 20 = 0. Solution:  JEE Main 20th July, 2021 Let r1 and r2 be the radii of the largest and smallest circles, respectively, which pass through the point (-4, 1) and having their centres on the circumference of the circle x2 + y2 + 2x + 4y - 4 = 0. If then a + b = ___ A 3 B 11 C 5 D 7  JEE Main 20th July, 2021 Let r1 and r2 be the radii of the largest and smallest circles, respectively, which pass through the point (-4, 1) and having their centres on the circumference of the circle x2 + y2 + 2x + 4y - 4 = 0. If then a + b = ___ A 3 B 11 C 5 D 7 Solution: Family of Circles  Family of Circles Just like family of lines, we have family of circles too, such as circles passing through intersection of two circles, a circle and a line etc.  Family of Circles (1) S + L = 0 S=0 L=0 E.g. Any circle passing through x2 + y2 − 2x −3 = 0 and x − 2y + 1 = 0 is of the form ____________. Find equation of circle passing through points of intersection of the line x + y − 1 = 0 and the circle x2 + y2 = 9 and which also passes through the point (3, 4). Solution:  Family of Circles (2) S + λS’ = 0, λ ≠ -1 S=0 S’ = 0  Family of Circles (2) S + λS’ = 0, λ ≠ -1 S=0 S’ = 0 NOTE S - S’ = 0 gives the equation of the common chord of S = 0 and S’ = 0.  JEE Main 2020 The circle passing through the intersection of the circles, x2 + y2 - 6x = 0 and x2 + y2 - 4y = 0, having its centre on the line, 2x - 3y + 12 = 0, also passes through the point A (-3, 6) B (-1, 3) C (-3, 1) D (1, -3)  JEE Main 2020 The circle passing through the intersection of the circles, x2 + y2 - 6x = 0 and x2 + y2 - 4y = 0, having its centre on the line, 2x - 3y + 12 = 0, also passes through the point A (-3, 6) B (-1, 3) C (-3, 1) D (1, -3) Solution: If the circle x2 + y2 + 4x + 22y + c = 0 bisects the circumference of the circle x2 + y2 − 2x + 18y − d = 0, then find c + d. Solution:  Family of Circles (3) Family of circles passing through two given points A(x1, y1) and B(x2, y2) A (x1, y1) B (x2, y2) E.g. Any circle through (1, 1) and (2, 2) is of the form  Family of Circles (3) Family of circles passing through two given points A(x1, y1) and B(x2, y2) Consider Then, S + 𝜆L = 0 gives family of circles passing through A(x1, y1) and B(x2, y2) E.g. Any circle through (1, 1) and (2, 2) is of the form (x − 1) (x − 2) + (y − 1) (y − 2) + 𝜆(y − x) = 0.  Family of Circles (4) Family of circles tangent to a given line at a given point L=0 A (x1, y1)  Family of Circles (4) Family of circles tangent to a given line at a given point L=0 A (x1, y1) Say that the given line and the point are respectively L = 0 and A (x1, y1). Consider, a circle through A (x1, y1), S : (x − x1)2 + (y − y1)2 Then S + 𝜆L = 0 gives a family of circles touching the line L = 0 at point A (x1, y1).  Family of Circles E.g. Any circle touching x + y + 1 = 0 at (1, − 2) is of the form (x − 1)2 + (y + 2)2 + 𝜆(x + y + 1) = 0. If the radius of the circle passing through the origin and touching the line x + y = 2 at (1, 1) is r units, then the value of is If the radius of the circle passing through the origin and touching the line x + y = 2 at (1, 1) is r units, then the value of is Ans: 3 Solution:  JEE Main 26th Aug, 2021 A circle C touches the line x = 2y at the point (2, 1) and intersects the circle C1 : x2 + y2 + 2y - 5 = 0 at two points P and Q such that PQ is a diameter of C1. Then the diameter of C is: A B 15 C D  JEE Main 26th Aug, 2021 A circle C touches the line x = 2y at the point (2, 1) and intersects the circle C1 : x2 + y2 + 2y - 5 = 0 at two points P and Q such that PQ is a diameter of C1. Then the diameter of C is: A B 15 C D Solution: Chords of Circles  Chords of Circles Here, we will be studying (1) Chord of contact (2) Chord with given midpoint  Chords of Circles (1) Equation of CoC (chord of contact) with respect to P(x1, y1) Its equation is given by T = 0 P (x1, y1) S=0 (2) Equation of chord with given midpoint P(x1, y1) Its equation given by T = S1 P (x1, y1) S=0 If the straight line x - 2y + 1 = 0 intersects the circle x2 + y2 = 25 in points P and Q, then find the coordinates of the point of intersection of tangents drawn at P and Q to the circle x2 + y2 = 25. Solution: x - 2y + 1 = 0 P R Q Tangents are drawn to the circle x2 + y2 = 16 at the points where it intersects the circle x2 + y2 - 6x - 8y - 8 = 0, then the point of intersection of these tangents is A (4, 16/3) B (12, 16) C (3, 4) D (16, 12) Tangents are drawn to the circle x2 + y2 = 16 at the points where it intersects the circle x2 + y2 - 6x - 8y - 8 = 0, then the point of intersection of these tangents is A (4, 16/3) B (12, 16) C (3, 4) D (16, 12) Solution: A C2 C1 P(x1 , y1) B