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Circle and Position of a Point wrt Circle - Part 1 for IIT JEE and other Exams

Posted on - 15-01-2017

JEE Math Circle

IIT JEE

Definition

A circle is the locus of a point which moves in such a way that its distance from a fixed point is constant. The fixed point is called the centre of the circle and the constant distance, the radius of the circle.

Equations of the Circle in Various Forms:

  • The simplest equation of the circle is x2 + y2 = r2 whose centre is (0, 0) and
    radius r.
  • The equation (x – a)2 + (y – b)2 = r2 represents a circle with centre (a, b) and
    radius r.
  • The equation x2 + y2 + 2gx + 2fy + c = 0 is the general equation of a circle with centre (-g, -f) and radius .
  • Equation of the circle with points P(x1, y1) and Q(x2, y2) as extremities of a diameter is (x – x1)(x – x2) + (y – y1)(y – y2) = 0.
  • The equation of the circle through three non-collinear points P(x1, y1) , Q(x2, y2) and R(x3, y3) is = 0.
  • Equation of a circle under different conditions
  • CONDITION
  • EQUATION
  • (i) Touches both the axes with centre (a, a)
  • and radius a
  • (x-a)2 + (y-a)2 = a2
  • (ii) Touches x-axis only with centre
  • (a, a) and radius a
  • (x-a)2 + (y-a)2 = a2
  • (iii)Touches y-axis only with centre (a, b) and
  • radius a
  • (x-a)2 + (y-b)2 = a2
  • (iv) Passes through the origin with centre
  • and radius .
  • x2 +y2 - a x - b x = 0
  • .
  • Parametric Equation of a Circle
  • The equations x = a cosq , y = a sinq are called parametric equations of the circle x2+ y2 = a2 and q is called a parameter. The point (a cos q, a sin q) is also referred to as point q. The parametric coordinates of any point on the circle (x – h)2+(y – k)2 = a2 are given by (h + a cosq, k + a sinq) with 0 £ q < 2p. .

    Example 1

    Find the centre and the radius of the circle 3x2+3y2-8x-10y+3 = 0.

    Solution:

    We rewrite the given equation as

    x2 + y2 - x - y + 1 = 0


    &⇒
    g = -, f = - , c = 1

    Hence the centre is and the radius is

    = =

    Example 2

    Find the equation of the circle whose centre is (1, 2) and which passes through the point (4, 6).

    Solution:

    The radius of the circle is = = 5.

    Hence the equation of the circle is (x – 1)2 + (y – 2)2 =25


    &⇒
    x2 + y2 – 2x – 4y = 20

    Example 3

    Find the equation of a circle passing through (1, 1), (2, -1) and (3, 2). .

    Solution:

    Let the equation be

    x2 + y2 + 2gx + 2fy + c = 0

    Substituting for the three points, we get

    2g + 2f + c = –2

    4g – 2f + c = –5

    6g + 4f + c = –13

    Solving the above three equations, we obtain:

    f = –1/2; g = –5/2 ; c = 4

    Hence the equation of the circle is

    x2 + y2 – 5x – y + 4 = 0.

    Example 4

    Find the equation of the circle whose diameter is the line joining the points (–4, 3) and (12, –1). Find also the intercept made by it on
    the y-axis.

    Solution:

    The equation of the required circle is

    (x + 4)(x – 12) + (y – 3) (y + 1) = 0

    On the y-axis, x = 0


    &⇒
    -48 + y2 – 2y – 3 = 0


    &⇒
    y2 – 2y – 51 = 0


    &⇒
    y = 1 ±

    Hence the intercept on the y-axis =

    Example 5

    A circle has radius equal to 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:

    Let the centre of the circle be (a, b). It lies on the line y = x – 1.


    &⇒
    b = a – 1
    . Hence the centre is (a, a – 1).


    &⇒
    The equation of the circle is

    (x – a)2 + (y – a + 1)2 = 9

    It passes through (7, 3)


    &⇒
    (7 – a)2 + (4 – a)2 = 9


    &⇒
    2a2 – 22a + 56 = 0


    &⇒
    a2 – 11a + 28 = 0


    &⇒
    (a – 4) (a – 7) = 0
    &⇒
    a = 4, 7.

    Hence the required equations are

    x2 + y2 – 8x – 6y + 16 = 0 and x2 + y2 – 14x – 12y + 76 = 0.

    The position of a point with respect to a circle :

    The point P(x1, y1) lies outside, on, or inside a circle S º x2 + y2 + 2gx + 2fy + c = 0, according as S1 º x12 + y12 + 2gx1 + 2fy1 + c > = or < 0.

     
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