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22) The given equ" of the circle (x+12) ² + (y + 5) ² = 169 Here centre of the circle ins (-12,-5) and radius is 13 unit. Now (x(+12) ² + (y + 5) = 169 9 Differentiate wir-t a we 2 ( x +¹²) + 2 (y + 5) dy = 0 वर → dy= y+5 • Em (60) = -¹² dy 12 S The equn of the (2-0) = - 1²/31 (^~0) x+12 =-122 sy 2x+ 5yco (Ang) 127- Pr (-17-5) tangent get at (0,0) 12n+8y=

22) The given equ" of the circle (x+12) ² + (y + 5) ² = 169 Here centre of the circle ins (-12,-5) and radius is 13 unit. Now (x(+12) ² + (y + 5) = 169 9 Differentiate wir-t a we 2 ( x +¹²) + 2 (y + 5) dy = 0 वर → dy= y+5 • Em (60) = -¹² dy 12 S The equn of the (2-0) = - 1²/31 (^~0) x+12 =-122 sy 2x+ 5yco (Ang) 127- Pr (-17-5) tangent get at (0,0) 12n+8y=

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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2) The (x+12)² + (7+5) ² = 169 Here centre of the circle in (-12,-5) and radius is 13 unit. given iven equ" of the circle is Now, (x+12) ² + (y + 5) * = 169 → Differentiate wir.t a we get 2(x + 2) + 2 (y + 5) dr = 0 क्य dy =. ㅋ ㅋ 127 En The equn of the (2-0) = - 1²/31 (x-0) 2+12 = =-122 y + 5 12 S tangent at (0,0) = sy: 127+ 5y =o (Ang) (-12-5) 124+3y=0
Transcribed Image Text:2) The (x+12)² + (7+5) ² = 169 Here centre of the circle in (-12,-5) and radius is 13 unit. given iven equ" of the circle is Now, (x+12) ² + (y + 5) * = 169 → Differentiate wir.t a we get 2(x + 2) + 2 (y + 5) dr = 0 क्य dy =. ㅋ ㅋ 127 En The equn of the (2-0) = - 1²/31 (x-0) 2+12 = =-122 y + 5 12 S tangent at (0,0) = sy: 127+ 5y =o (Ang) (-12-5) 124+3y=0
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