Question

Find the area of the triangle formed by the normal at (3, -4) to the circle x2+y2–22x–4y+25 = 0
625
with the coordinate axes.
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Answer 1

Given :  x²+y²–22x–4y+25 = 0

To Find :  area of the triangle formed by the normal at (3, -4) to the circle

Solution:

x²+y²–22x–4y+25 = 0

=> 2x  + 2ydy/dx  - 22  - 4dy/dx = 0

=>  x  +  ydy/dx  - 11 - 2dy/dx = 0

=> (dy/dx)(y - 2)  = 11 - x

=> dy/dx = (11 - x)/(y - 2)

Slope of tangent at  ( 3 , - 4)

 dy/dx = (11 - 3)/(-4 - 2)  = 8/(-6)  =  - 4/3

Slope of normal =  3/4

Equation of line passing through ( 3 , - 4)

y  -(-4)  = (3/4)(x - 3)

=> 4y  + 16   = 3x  -  9

=> 3x - 4y = 25

x = 0   => y  = -25/4

y = 0  => x  = 25/3

Area of triangle formed = (1/2)(25/4)(25/3)

= 625/24

= 26.04  sq unit

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