Question

Find the equation of the circle passing through (6,-5) and
having centre at (3,-1).​

Answer 1

Answer:

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Answer 2

Answer:

(a) Let required equation of the circle is x2 + y2 + 2gx + 2fy + c = 0 It passes through (0,5) & (6, 1) (0,5) 0 + 52 + 2g(0) + 2f(5) + c = 0 1f + C + 25 = 0 …(1) (5, 1) 62 + 12 + 2g (6) + 2f (1) + c = 0 12g + 2f + c + 37 = 0 …..(2) Centre (-g, -f) lies on the line 12x + 5y – 25 = 0 -12g – 5f – 25 = 0 ….(3) Eqn. 2 – eqn. (1) gives 12g – 8f + 12 = 0 …..(4) Adding 3 & 4 we get 12g – 13f – 13 = 0 ⇒ f = -1 From eqn (1) ⇒ 10(-1) + C + 25 = 0 ⇒ C = – 15 ∴ -12g = 5f + 25 -12g = 20 ⇒ g = 20−1220−12= −53−53 ∴ The required equation of the circle is x2 + y2 + 2 (−53)(−53)x + 2(-1)y – 15 = 0 ⇒ × 3 3x2 + 3y2 – 10x – 6y – 45 = 0 (b) Let the required equation of the circle is x2 + y2 + 2gx + 2fy + c = 0 but it passes through (1,-4) & (5,2) & centre (-g, -f) lies on the line ⇒ -g + 2f + 9 = 0 …… (1) (1,-4) 12 + (-4)2 + 2g(1) + 2f(-4) + c = 0 2g – 8f + C + 17 = 0 ….. (2) ⇒ (5,2)52 + 22 + 2g(5) + 2f(2) +C = 0 10g + 4f + C+ 29 = 0 ….. (3) Eqn. (3) – eqn. (2) gives 8g + 12f + 12 = () – 4 2g + 3f + 3 = 0 …..(4) 2f + 9 = g ⇒ g = -6 + 9 = 3 C = -2g + 8f – 17 C = -6 -24 – 17 = – 47 ∴ The required equation of the circle is x2 + y2 + 2 (3)x + 2(-3)y – 47 = 0 x2 + y2 + 6x – 6y – 47 = 0 (c) Let the equation of required circle is x2 + y2 + 2gx + 2fy + c = 0 it passes Through (0, -3) & (0,5) & centre (-g, -f) Lies on the line x – 2y + 5 = 0 ⇒ -g + 2f + 5 = 0 …..(1) (0,-3) 0 + 9 + (-6f) + c = 0 -6f + 9 + c = 0 …..(2) (0,5) 0 + 25 + 0 + 10f + C = 0 10f + C + 25 = 0 …. (3) ∴ eqn. (3) – eqn. (2) gives 16f + 16 = 0 ⇒ f =-1 2f + 5 = g ⇒ g= -2 + 5 = 3 c = 6f – 9 = -6 -9 = -15 ⇒ c = -15 ∴ The equation of the circle with g = 3, f = -1 & C = -15 is x2 + y2 + 2(3)x + 2(-1)y – 15 = 0 x2 + y2 + 6x – 2y – 15 = 0 (d) Let the equation of the required circle is x2 + y2 + 2gx + 2fy + c = 0 If passes through (1, 1) & (2,2) (1,1)12 + 12 + 2g(1) + 2f(1) + c = 0 2g + 2f + C + 2 = 0 ….. (1) (2, 2)4 + 4 + 2g(2) + 2f(2) + c = 0 4g + 4f + c + 8 = 0 …… (2) Also r = 1 ⇒ g2+f2−c−−−−−−−−−√g2+f2−c =1 ⇒ g2 + f2 – c = 1 …… (3) Eqn. (2) – eqn. (1) gives 2g + 2f + 6 = 0 g + f + 3 = 0 g = (-3 -f) ….(4) Adding 1 and 3 we get g2 + f2 + 2g + 2f = -1 (-3 -f)2 + f2 + 2 (-3-f) + 2f + 1 = 0 f2 + 9 + f2 + 6 - 6 - 2f + 2f + 1 = 0 2f2 + 6f + 4 = 0 ÷ 2 f2 + 3f + 2 = 0 (f + 2) (f + 1) = 0 ⇒ f = -1 or -2 When f = -2, g = -3 - f = -3 + 2 = -1 When f = -1, g = -3 - f = -3 + 1 = -2 C = g2+ f2 -1 = 1 + 4 – 1 = 4 ∴ Required circles are x2 + y2 – 2x – 4y + 4 = 0 and x2 + y2 – 4x – 2y + 4 = 0.