Question

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Answer 1

HI :DD

Answer 2

Differentiate the equation (with respect to x)

Answer 3

Can you tell me, what will it give after differentiating the equation x^2 + y^2 = 169 w.r.t x?

Answer 4

Okay, let us first differentiate x^2 w.r.t x

Do you know the basics of differentiation?

Answer 5

putting the co ordinates of point in the equation of circle u will find that point is on the circle
therefore tangent to a circle. = t=0
t=x(x1) + y(y1)=0 (when center is at origin)
5x-12y=169
now slope = - a/b
=5/12

Answer 6

-[ 5/(-12)]

Answer 7

a=coef. of x
b= coef of y

Answer 8

That can be one of the methods to solve too. My method was :

\(\boxed{\textbf{Differentiating the equation w.r.t x, we get :} \\
\mathsf{ 2x + 2y (\dfrac{dy}{dx}) = 0 } \\
\textbf{Since , } \space \mathsf{\dfrac{dy}{dx}} \space \textbf{is the slope of the line that is tangent to the given circle, so } \\
\textbf{, let it be m.} \\
\mathsf{ 2x + 2ym = 0 \\
m = \dfrac{-2x}{2y} = \dfrac{-x}{y} \\
\implies m = -(\dfrac{-5}{12}) = \dfrac{5}{12} \\ }}\)

Answer 9

@vachave412 - Where are you confused ? @nirmalnema 's method was easier , let us know where are you stuck? And we can try to help you out.

Answer 10

Oh okay , fine! :) Good Luck.