Question

Equation of tangent corresponding to the normal whose equation is y = x to the circle x2 + y2 – 4x – 4y + 4 = 0 is

Answer 1

Given : normal  equation is y = x to the circle x2 + y2 – 4x – 4y + 4 = 0

To find : Equation of tangent corresponding to the normal

Solution:

x² + y² – 4x – 4y + 4 = 0

y = x is the normal

Point is

x² + x² - 4x - 4x + 4 = 0

=> x² - 4x + 2 = 0

=> x  = (4± √16-8)/2

=> x =  2 ± √2  

y = x

Point are ( 2 ± √2  , 2 ± √2)

normal y = x

Slope of normal = 1

Slope of tangent = - 1

y - (2 ± √2) = - (x -  (2 ± √2)

=> y   =  -x +  2(2 ± √2)

=> x + y =  4  ± 2√2

x + y =  4  ± 2√2  is the Equation of the tangent  corresponding to the normal whose equation is y = x  

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