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1aCircles Part I

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CirclesI
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   www.sakshieducation.com www.sakshieducation.com CIRCLES PART - I Equation of A Circle: The equation of the circle with centre C (h, k) and radius r is (x – h) 2  + (y – k) 2  = r 2 . Proof:  Let P(x 1 , y 1 ) be a point on the circle. P lies in the circle ⇔  PC = r ⇔ 2211 (xh)(yk)r − + − =   ⇔  (x 1  – h) 2  + (y 1  – k) 2  = r 2 . The locus of P is (x – h) 2  + (y – k) 2  = r 2 . ∴  The equation of the circle is (x–h) 2  + (y–k) 2  = r 2 .------(1) Note: The equation of a circle with centre srcin and radius r is (x–0) 2  + (y–0) 2  = r 2 i.e., x 2  + y 2  = r 2  which is the standard equation of the circle. Note: On expanding equation (1), the equation of a circle is of the form x 2  + y 2  + 2gx + 2fy + c = 0. Theorem: If g 2  + f  2  – c ≥≥≥≥  0, then the equation x 2  + y 2  + 2gx + 2fy + c = 0 represents a circle with centre (–g, –f) and radius 22 gfc + − . Note:  If ax 2  + ay 2  + 2gx + 2fy + c = 0 represents a circle, then its centre = gf ,aa   − −     and its radius 22 gfac|a| + − .   www.sakshieducation.com www.sakshieducation.com Theorem: The equation of a circle having the line segment joining A(x 1 , y 1 ) and B(x 2 , y 2 ) as diameter is 1212 (xx)(xx)(yy)(yy)0 − − + − − = . Let P(x,y) be any point on the circle. Given points A(x 1 , y 1 ) and B(x 2 , y 2 ). Now 2  APB  π   = . (Angle in a semi circle.) Slope of AP. Slope of BP =-1 ( )( ) ( )( )( )( ) ( )( ) 121221212121 10  y y y y x x x x y y y y x x x x x x x x y y y y − − ⇒  = −− − ⇒  − − = − − − ⇒  − − + − − =   Definition:  Two circles are said to be concentric if they have same center. The equation of the circle concentric with the circle x 2  + y 2  + 2gx + 2fy + c = 0 is of the form x 2  + y 2  + 2gx + 2fy + k = 0. The equation of the concentric circles differs by constant only. Parametric Equations of A Circle: Theorem: If P(x, y) is a point on the circle with centre C( α , β ) and radius r, then x = α  + r cos θ , y = β  + r sin θ  where 0 ≤θ < 2 π .   www.sakshieducation.com www.sakshieducation.com Note: The equations x = α  + r cos θ , y = β  + r sin θ , 0 ≤θ < 2 π  are called parametric equations of the circle with centre ( α , β ) and radius r. Note: A point on the circle x 2  + y 2  = r 2  is taken in the form (r cos θ , r sin θ ). The point (r cos θ , r sin θ ) is simply denoted as point θ . Theorem:   (1) If g 2  -c > 0 then the intercept made on the x axis by the circle x 2  + y 2  + 2gx + 2fy + c = 0 is 2 2  g ac −  2) If f  2  – c >0 then the intercept made on the y axis by the circle x 2  + y 2  + 2gx + 2fy + c = 0 is 2 2  f bc −   Note: The condition for the x-axis to touch the circle x 2  + y 2  + 2gx + 2fy + c = 0 (c > 0) is g 2  = c. Note:  The condition of the y-axis to touch the circle x 2  + y 2  + 2gx + 2fy + c = 0 (c > 0) is f  2  = c. Note: The condition for the circle x 2  + y 2  + 2gx + 2fy + c = 0 to touch the coordinate axes is g 2  = f  2  = c. Position of Point: Let S = 0 be a circle and P(x 1 , y 1 ) be a point I in the plane of the circle. Then i) P lies inside the circle S = 0 ⇔  S 11 < 0 ii) P lies in the circle S = 0 ⇔  S 11  = 0 iii) P lies outside the circle S = 0 ⇔  S 11  = 0 Power of a Point: Let S = 0 be a circle with centre C and radius r. Let P be a point. Then CP 2  – r 2  is called power of P with respect to the circle S = 0.   www.sakshieducation.com www.sakshieducation.com Theorem: The power of a point P(x 1 , y 1 ) with respect to the circle S = 0 is S 11 . Theorem:  The length of the tangent drawn from an external point P(x 1 , y 1 ) to the circle S = 0 is 11 S . Very Short Answer Questions   1. Find the equation of the circle with centre C and radius r where. i) C = (1, 7), r = Sol.  Equation of the circle is ( x-h) 2  + (y-k) 2  = 2   ⇒ ( x-1) 2  + (y-7) 2  = 2   ⇒ x 2  – 2x + 1 + y 2  – 14y + 49 = ⇒ x 2  + y 2  – 2x – 14y + 0 ⇒  4x 2  + 4y 2  – 8x – 56y + 175 = 0 ii) C = (a, -b); r = a + b Equation of the circle is (x-h) 2  + (y-k) 2  = 2  Equation of the circle is ( x-a) 2  + (y-(-b)) 2  = 2   ⇒  x 2  – 2xa + a 2  + y 2  +2by + b 2 = a 2  +2ab + b 2   ⇒  x 2  + y 2  – 2xa + 2by – 2ab = 0 2. Find the equation of the circle passing through the srcin and having the centre at (-4, -3). Sol . Centre (h, k) = (-4, -3) Equation of the circle is (x – h) 2 + ( y –k) 2  = r 2 ; (x +4) 2  + (y +3) 2  = r 2  Circle is passing through srcin ∴ (0 +4) 2  + (0 + 3) 2  = r 2
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