Circle and Conic Section

Answer :

circle = x² + y² – 2ax – ay + a² = 0

                        ∴ centre: [a, (a/2)]       radius = √[a² + (a² / 4) – a²] = (a/2)

                        Here radius = y co-ordinate of centre

                        ∴ Radius = distance of centre from x axis.

                        ⇒ Circle touches x axis.         

Answer :

Equation: [{m(x – 1)²} / 3] + [(y – 1)² / 4] = 1

                        ∵ It represents a circle

                        ∴ Coefficient of x² = coefficient of y²                                                           (1)

                        Equation: (m/3)(x² – 2x + 1) + (1/4)(y² – 2y + 1) = 1

                        ∴ from (1), (m/3) = (1/4) ⇒ m = (3/4).           

Answer :

Circle: x² + y² – 2x + 4y – 93 = 0

                        Centre: (1, – 2)

                        Radius = √(1² + 2² + 93) = √(98)

                        Length of square = diameter of circle. Let side of square is a,

                        ∴         √(a² + a²) = 2 × √(98)

                        ∴         2a² = 98 × 4

                        ∴         a² = 196

                        ∴         a = 14

                        ∵ Sides of square are parallel to the co-ordinate axes,

                        ∴ Distance of each side from centre is 7.

                        ∴ sides parallel to y axis are x = 1 + 7 and x = 1 – 7

                        ∴         x = 8 and x = – 6

                        also sides parallel to x axis are y = – 2 + 7 = 5 and y = – 2 – 7 = – 9.

                        ∴ Vertices of square are (– 6, – 9)(– 6, 5)(8, – 9)(8, 5)

Answer :

                        Circle: x² + y² – 4x – 4y + 4 = 0                                                                     (1)

                        ∴ Centre is (2, 2) and radius = 2

                        in ΔAOB,

                        Let equation of AB be (x/a) + (y/b) = 1

                        ∴         A(a, 0) and B(0, b)

                        AB is diameter of circumcircle of ΔAOB.

                        ∴ It’s centre say M(p, q) is midpoint of AB.

                        ∴         M = [{(a + 0) / 2} and {(0 + b) / 2}]

                        ∴         P = (a/2) and q = (b/2) ⇒ a = 2p and b = 2q

                        ∵ AB touches circle in equation (1), ⇒ r = distance of point (2, 2) from line

                        (x/a) + (y/b) = 1.

                                    |r| = 2 = |[{(2/a) + (2/b) – 1} / √{(1 / a²) + (1 / b²)}]|

                        ∴         2a + 2b – ab = ± 2√(a² + b²)

                        i.e.       4p + 4q – 4pq = ± 2√(4p² + 4q²) ---------- a = 2p and b = 2q

                        ∴ locus of circumcenter is 4x + 4y – 4xy = ± 4√(x² + y²)

                        i.e.       x + y – xy ± √(x² + y²) = 0.

                        ∴ by comparison, we have k = 1.      

Answer :

Answer:           (c) 75 sq. units

                        The centre of circle is (1, 2) equation of tangent at A & B are

                                    x(1) + y(7) – (x + 1) – 2(y + 7) – 20 = 0                                             (1)

                        and

                                    4x – 2y – (x + 4) – 2(y – 2) – 20 = 0                                       (2)

                        i.e.       x + 7y – x – 1 – 2y – 14 – 20 = 0        from    (1)

                        ⇒         5y = 35

                        i.e.       y = 7

                        and

                                    3x – 4y – 20 = 0 ------- from (2)

                        ⇒         3x = 20 + 4y = 48

                                    x = 16

                        ∴ point D(16, 7)

                        ∴ area of quadrilateral

                                    ABCD = 2 × area of ACD

                                                = |(7 – 7) – 2(1 – 16) + 1(7 – 112)|

                                                = |– 75|

                                                = 75 sq. units.