Question

Equation of normal drawn from point (6 8) to circle $x^{2} +y^{2} -4x-6y+3=0$

Answer 1

Given is the equation of circle ,

x^{2} + y^{2} - 4x- 6y + 3 = 0

Comparing with the general formula of circle , we get ,

f = -2 , g = -3

Hence the radius is (-2,-3)

Substituting in the formula of Equation of Normal ,

y - y1 = [(y1+f)/(x1+g)] (x-x1)

Applying the coordinates of radius,

y - 3 = 5/4(x - 2)

Hence 5x - 4y + 2 = 0 is the equation of normal.

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Answer 2

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