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01_Straight_Line 02_Circle 03_Permutation_and_combination 04_Complex_numbers 05_Application_of_Derivative 06_Binomial_theorem 07_Progression__Series.pdf 08_Definite_Integral.pdf 09_Indefinite_Integral.pdf 10_Probability.pdf

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    CIRCLE DEFINITIONS A circle is the locus of a point which moves in such a way that its distance from a fixed point, called the centre, is always a constant. The distance r from the centre is called the radius of the circle. Twice the radius is known as the diameter d = 2r. The perimeter C of a circle is called the circumference, and is given by C = ltd = 2Ttr. The angle a circle subtends from its centre is a full angle equal to 3600 or 211 radians. Secant: A line, which intersects a circle in two distinct points, is called a secant. Tangent: A line meeting a circle only in one point is called a tangent to the circle at that point. The point at which the tangent line meets the circle is called the point of contact. Length of tangent: The length of the line segment of the tangent between a given point and the given point of contact with the circle is called the length of the tangent from the point to the circle. Equation of a circle o Secant Tangent The equation of a circle with (h, k) as its centre and a with radius is (x — + (y — = a2 when the centre is at the origin, the equation of the circle become + Y2 = a . 2 General equation of a circle is + + 2gx + 2fy + c = 0 or (x + + (y + = g2 + f2 — c. So that its centre is at (— g, —f) and its radius is g2 +f2 —c . The equation of the circle with (Xl, yo and Y2) as the extremities of arc of its diameter is Example 1 : Solution : If the equations of the two diameters of a circle are x + y = 6 and x + 2y = 4 and the radius of the circle is 10, find the equation of the circle. Here radius of the circle = 10. Equations of two diameters say AB and ML of the circle are respectively and solving (i) and (ii), we get x = 8 and y = -2. Hence centre of the circle is (8, -2). Now the equation of the required circle is 102 or +4y -32 = O. Position of a Point The point P(XI, yo lies inside the circle S = + + 2gx + 2fy + c = 0 if S Xl + y 1 + + 2fY1 + c < . The point P(XI, yo lies on the circle S = 0 if S The point P(XI, yo lies outside the circle S = 0 if S Example 2 : Solution: How are the points (0, 1), (3, 1) and (1 , 3) situated with respect to the circle Given equation is + - 2x - 4y +3 = 0 Let (0, 1), (3, 1) and C = (1, 3) For point A(O, 1), + - 2x - 4y +3 = 02 + 12 Hence point A lies on the circle For point B(3, 1), + - 2x - 4y +3=3 > 0 Hence, point B lies outside the circle. For point C(l, 3), + - 2x - 4y Hence point C lies inside the circle. 2.0 4.1 +3=0
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    2 Two circles Sl x y 2 + 2f1Y + Cl = O and Circle S2 = x + y + + 2f2Y + = 0 be two circles with centre at A(— gl, —fl) and B(— g2, — f2) and radii rl and r2. Circle Sl = 0 is outside the circle S2 = 0 (and the circles are non-intersecting), then AB > rl + r2. If the two circles are touching externally, then AB = rl + r2. If the two circles are touching internally, then AB = Iri — If the circle Sl = 0 is completely inside the circle S2 = 0, then AB < Iri— r21. Example 3 : Prove that the circle + + 2ax + c 2 = 0 and + + 2by + c 2 1 1 1 Solution: Given circles are and + + 2by + c2 = 0 = 0 touch each other if Let A and B be the centres of circles (i) and (ii) respectively and rl and r2 be their radii, then The two circles (i) and (ii) will touch each other externally or internally according as AB = + or AB = Irl i.e., AB2 = (rl + r2)2 or AB2 = (rl 2 Thus the two circles will touch each other if AB2 = (r I ± r 2) 2 or a2 + b2 or a2 + b2 = a2 -c2 + b or 2c2 a2 —c2 b2 orc4 = (a 1 1 A Line and a Circle c2) (b2 - c2) 1 C c2±2 a2 -c2 b2 -c2 2 C or or c4 = a2b2 - c2b2 [dividing by a2b2c2] 4 or 2 c c2b2 + a2c2 = a2b2 The line ax + by + c = 0 and the circle S x + y + 2gx + 2fy + c = 0 intersect in two points whose coordinates are obtained by solving the equations L = 0, S = 0 simultaneously. The resulting equation, after eliminating x (or y) is a quadratic equation which may have real (distinct or equal) or imaginary roots. If the roots are distinct, the points of intersection are real, if the roots are equal , then line touches the circle. In case of imaginary roots, the line does not cut or touch the circle. Example 4 : Solution: Show that the line 3x - 4y - c = 0 will meet the circle having centre at (2, 4) and the radius 5 in real and distinct points if -35 < c < 15. Given line is 3x - 4y -c = 0 Centre of given circle is (2, 4) and its radius is 5, therefore its equation will be or From (i), y 1 4 + —(3x 16 = 52 8y-5=O c). Putting the value of y in (ii), we get -4x-8.-(3x-c) or or 16x2 + - 6cx + c2 - - + 32c - 80 = o - 2(80 + 3c) x + c2 + 32c - 80 = o ...(iii) Line (i) will meet the circle (ii) in real and distinct points if discriminant of equation (iii) > 0 i.e., if + 3c)2 100 (c2 + 32c - 80) > o or (80 + 3c)2 - 25(c2 + 32c - 80) or or or 6400 + 9c2 + - 25c2 - 800c + 2000 > o 16c2 - + 8400 or 16c2 + 320c- 8400 < o c2 + 20c- 525< o
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    Circle sign scheme for c2 + 20c — 525 : When c2 + 20c - 525 = O, -20+ 400+2100 C 2 3 -20+50 35,15 2 Therefore, sign scheme for c2 + 20c - 525 is as follows -35 c2 + 20c - 525 < O -35 < c < 15. Tangent and Normal to a Circle The equation of the tangent to the circle S = 0 at (Xl, yo is The equation of the tangent to the circle + Y2 = a2 at (Xl, yo is 2 + yyl = a . 15 The line(s) y = m. Example 5 : Solution: mx ± a 1+m2 is a tangent to the circle (or touches the circle) + = a2 for all real values of Find the equation of the normal to the circle + y 2 = 0 at the point 2x - 4y + 3 Given circle is + - 2x - 4y + 3 = 0 Let P = (2, 3) Let A be the centre of circle (i), then A (1 , 2) Normal at point P(2, 3) of the circle will be line AP. or The lines 2-3 1-2 y-3=x-2 or x -y +1 = 0. f 2 +g2 —c , is tangent to the circle + + 2gx + 2fy + c = 0 for all m. The line y— b = m(x — a) ± touches the circle (x — + (y — b) 2 = r2 for all m. The line L = 0 touches the circle S = 0 if the perpendicular from the centre of the circle to the line L = 0 IS equal to the radius of the circle. x 12 -k y 12 + c The length of the tangent to the circle S = 0 from a point P(XI, yo is S From a point P(XI, yo outside the circle S = 0, two tangents can be drawn to the circle. The equation of the pair of tangents from P(XI, yo to S = 0 is SSI = T2 which Sl = S(XI, yo and T = + + g(x + + f(y + yo + c. Any line passing through the centre of the (x — xo is normal to circle is normal to the circle at the point where it meets the circle. The line y — = the circle S = 0 at the point (Xl, yo. Parametric Equation of a Circle Any point on the circle + = a2 may be written as (a cos 0, a sin 0). x = a cos 0, y = a sin 0 are the parametric equations of a circle. The equation of the tangent to the circle at any point is x cos 0 + y sin 0 = a. Any point in the circle (x — + (y — = r2 is (a + r cos 0, B + r sin 0). Chords of a Circle From a point P(XI, yo let us draw two tangents to the circle S = 0 so as to touch it at Q and R. The line QR is called the chord of contact of the two tangents to the circle S = 0. The equation of the chord of contact is + + g(x + + f(y + b) + c = 0. If the circle is + Y2 = a2, then the corresponding equation of the chord of contact is + yyl = a . The chord of the circle S = 0 whose midpoint is (Xl, yo is Sl = T i.e. S(XI, yo = T or, Xl + y 1 4-29(1 + c = + + f (y -k + c . The equation of the chord of the + = a2 with mid point (Xl, yo is + yi = + yyl .
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    4 Radical Axis of Two Circles Circle Let Sl = 0 and S2 = 0 be two circles with the coefficients of and Y2 equal to unity. Let the two circles be non- intersecting the locus of the point P, such that the length of the tangents to the two circle from it are equal, is called the radical axis of the two circles. Equation of the radical axis of Sl = 0, S2 = 0 is Sl — S2 = 0, which is linear in and y and hence represents a straight line. When the circles are intersecting each other, the radical axis becomes the common chord of the two circles with its equation Sl — S2 = 0. When the circles are touching each other (internally or externally) then the common chord becomes the common tangent to the two circles, with its equation Sl — S2 = 0. Orthogonal Circles Let SI = 0 and S2 = 0 be two circles with radii rl and r2 respectively. The two circles are said to cut orthogonally if the tangents to the two circles at their point of intersection are perpendicular to each other. It means that AB2 = r12 +r: (where A, B are the centres of the circles) 2f1f2 + 2g1g2 = Cl + c2. Example 6 : Solution: Obtain the equation of the circle orthogonal to both the circles + Y2 + 3x 4x2 + 4y2 - 28x + 29 = 0 and whose centre lies on the line 3x + 4y + 1 = 0. Given circles are + + 3x- 5y + 6 = 0 and +4y2 28x + 29 = O 29 0. 4 - 0 and 5y + 6 — Let the required circle be + + 2gx + 2fy + c = 0 Since circle (iii) cuts circles (i) and (ii) orthogonally 3 5 2 2 and 2g 7 or 7g 29 + 2f.O = c + 2 4 4 (iv) - (v), we get 10g - 5f = 5 — or 40g - 20f = 4 5. (Ill) ... (iv) ...(v) ... (vi) ... (vii) (viii) Given line is 3x + 4y + 1 = 0 Since centre (-g, -f) of circle (iii) lies on line (vii) 3g - 4f = 1 1 Solving (vi) and (viii), we get g = 0 and f = — 4 29 . from (v), c = 4 . From (iii), required circle is 1 29 O or 4(x2 + y2) +2y-29 = O. 4 Family of Circles From two points, an infinity of circles can be made to pass. All these circles belong a family. The equation of the family of circles passing through the points of intersection of the line L = 0 and the circle S = 0 is S + Xl- = 0 where is an arbitrary constant whose value can be obtained from given geometrical conditions to fix the circle. The equation of the family of circles passing through the points of intersection of the circles Sl = 0 and S2 = 0 is + = O. The equation of the family of circles touching a given line L = ax + by + c = 0 at the given point (Xl, yo is (x — The equation of the family of line passing through two given points (Xl, yo and (x2, Y2) is X Y2 1
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    Circle Example 7 : Find the equation of the circle described on the chord x cos + y sin u —p = 0 of + y —a2 = 0 as diameter. Solution: Any such circle is a member of + —a2 + k (x cos + y sin u — p) = 0. If this is to be the circle on the chord as diameter, the centre of the circle —a 5 k — cosu, 2 k — sinu 2 should lie on x cos u + y sin u —p = 0 k — COS2 + 2 k Sin2 01 = p k = -2p. 2 The equation to the required circle is + y — 2P (x cos u + y sin u — p) = 0.

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