Class XI Chapter **11** – Conic Section Maths ______________________________________________________________________________ 2 2 2h 3 25 2h 25 9 2h 16 2 2 2h 16 4 If 2-h=4, then h=-2. If 2-h=-4, then, h=6. When h = â€“2, the equation of the circle becomes (x + 2) 2 + y 2 = 25 X 2 + 4x + 4 + y 2 = 25 X 2 + y 2 + 4x â€“ 21 = 0 When h = 6, the equation of the circle becomes (x â€“ 6) 2 + y 2 = 25 X 2 â€“ 12x +36 + y 2 = 25 X 2 + y 2 â€“ 12x + **11** = 0 Question 13: Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes. Solution 13: Let the equation of the required circle be (x â€“ h) 2 + (y â€“ k) 2 = r 2 . Since the circle passes through (0, 0), (0 â€“ h) 2 + (0 â€“ k) 2 = r 2 ⇒ h 2 + k 2 = r 2 The equation of the circle now becomes (x â€“ h) 2 + (y â€“ k) 2 = h 2 + k 2 . It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore, (a â€“ h) 2 + (0 â€“ k) 2 = h 2 + k 2 … (1) (0 â€“ h) 2 + (b â€“ k) 2 = h 2 + k 2 … (2) From equation (1), we obtain a 2 â€“ 2ah + h 2 + k 2 = h 2 + k 2 ⇒ a 2 â€“ 2ah = 0 ⇒ a(a â€“ 2h) = 0 ⇒ a = 0 or (a â€“ 2h) = 0 However, a ≠ 0; hence, (a â€“ 2h) = 0 ⇒ h = 2 a . 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 ______________________________________________________________________________ From equation (2), we obtain h 2 + b 2 â€’’ 2bk + k 2 = h 2 + k 2 ⇒ b 2 â€“2bk = 0 ⇒ b(b â€“ 2k) = 0 ⇒ b = 0 or (b â€“ 2k) = 0 However, b ≠ 0; hence, (b â€“2k) = 0 ⇒ k = 2 b . Thus, the equation of the required circle is 2 2 2 2 a b a b x y 2 2 2 2 2 2 2 2 2x a 2y b a b 2 2 4 4x 4ax a 4y 4by b a b 2 2 2 2 2 2 2 2 4x 4y 4ax 4by 0 2 2 x y ax by 0 Question 14: Find the equation of a circle with centre (2, 2) and passes through the point (4, 5). Solution 14: The centre of the circle is given as (h, k) = (2, 2). Since the circle passes through point (4, 5), the radius (r) of the circle is the distance between the points (2, 2) and (4, 5). 2 2 2 2 r 24 25 2 3 49 13 Thus, the equation of the circle is x h y k r 2 2 2 2 2 x 2 y 2 13 2 2 x x y y 4 4 4 4 13 2 2 x y x y 4 4 5 0 2 Printed from Vedantu.com. Register now to book a Free LIVE Online trial session with a Top tutor.

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