Class XI Chapter **11** – Conic Section Maths ______________________________________________________________________________ x a 2 y b 2 a 2 b 2 2 x 2ax a y 2by b a b 2 2 2 2 2 2 2 2 2 x y 2ax 2by 2b 0 Question 6: Find the centre and radius of the circle (x + 5) 2 + (y - 3) 2 = 36 Solution 6: The equation of the given circle is (x + 5) 2 + (y - 3) 2 = 36. (x + 5) 2 + (y - 3) 2 = 36 ⇒ {x - (-5)} 2 + (y - 3) 2 = 6 2 , which is of the form (x - h) 2 + (y - k) 2 = r 2 , where h = -5, k = 3, and r = 6. Thus, the centre of the given circle is (-5, 3), while its radius is 6. Question 7: Find the centre and radius of the circle x 2 + y 2 - 4x - 8y - 45 = 0 Solution 7: The equation of the given circle is x 2 + y 2 â€“ 4x â€“ 8y â€“ 45 = 0. X 2 + y 2 â€“ 4x â€“ 8y â€“ 45 = 0 ⇒ (x 2 â€“ 4x) + (y 2 â€“ 8y) = 45 ⇒ {x 2 â€“ 2(x)(2) + 2 2 } + {y 2 â€“ 2(y)(4)+ 4 2 } â€“ 4 â€“16 = 45 ⇒ (x â€“ 2) 2 + (y â€“4) 2 = 65 ⇒ (x â€“ 2) 2 + (y â€“4) 2 = 2 65 , which is of the form (x â€“ h) 2 + (y â€“ k) 2 = r 2 , where h = 2, k = 4, and r = 65 Thus, the centre of the given circle is (2, 4), while its radius is 65 . Question 8: Find the centre and radius of the circle x 2 + y 2 - 8x + 10y - 12 = 0 Solution 8: The equation of the given circle is x 2 + y 2 â€“ 8x + 10y â€“ 12 = 0. X 2 + y 2 â€“ 8x + 10y â€“ 12 = 0 Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

Class XI Chapter **11** – Conic Section Maths ______________________________________________________________________________ ⇒ (x 2 â€“ 8x) + (y 2 + 10y) = 12 ⇒ {x 2 â€“ 2(x)(4) + 4 2 } + {y 2 + 2(y)(5) + 5 2 }â€“ 16 â€“ 25 = 12 ⇒ (x â€“ 4) 2 + (y + 5) 2 = 53 2 2 2 x y 4 5 53 , which is of the form (x â€“ h) 2 + (y â€“ k) 2 = r 2 , where h = 4, k = â€“5, and r = 53 . Thus, the centre of the given circle is (4, â€“5), while its radius is 53 Question 9: Find the centre and radius of the circle 2x 2 + 2y 2 - x = 0 Solution 9: The equation of the given circle is 2x 2 + 2y 2 â€“ x = 0. 2 2 2x 2y x 0 2 2 2x x 2y 0 2 x 2 2x 0 2 y 2 2 2 1 1 2 1 x 2. x y 0 4 4 4 2 2 2 1 1 x y0 4 4 , which is of the form (x â€“ h)2 + (y â€“ k) 2 = r 2 , where h = 1 4 , k = 0, and r = 1 4 Thus, the centre of the given circle is ( 1 4 , 0) , while its radius is 1 4 Question 10: Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16. Solution 10: Let the equation of the required circle be (x â€“ h) 2 + (y â€“ k) 2 = r 2 . Since the circle passes through points (4, 1) and (6, 5), (4 â€“ h) 2 + (1 â€“ k) 2 = r 2 … (1) (6 â€“ h) 2 + (5 â€“ k) 2 = r 2 … (2) Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

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