Question

find the equation of the tangent to the circle x^2+y^2=25 drawn through the point(13,0).

Answer 1

\[\frac{dy}{dx}=-\frac{x}{y}\] so the slope between the point \((x,y)\) on the circle and \((0,13)\) must satisfy the equation
\[\frac{y-13}{x}=-\frac{x}{y}\]
and since it is on the circle they must also satisfy the equation \(x^2+y^2=25\)

Answer 2

solve the two equations to get your answer

Answer 3

well i got answer 12y=+ and -5(x-13) is it true?

Answer 4

oh damn it is \((13,0)\) not \((0,13)\) ok idea is still the same but equation is wrong
\[\frac{y}{x-13}=-\frac{x}{y}\]

Answer 5

well i still didn't get it what u r trying to say

Answer 6

maybe i am mistaken.
you want the equation of the line tangent to the circle
i took the derivative of \(x^2+y^2=25\) with respect to \(x\) to find the equation for the slope.
are you using a different method?

Answer 7

yes i guess so

Answer 8

i just checked and your answer looks correct

Answer 9

just for my curiosity, what method are you using?

Answer 10

whatever you did, it worked
nice picture here
http://www.wolframalpha.com/input/?i=12y%3D5%28x-13%29%2C+x^2%2By^2%3D25%2C12y%3D-5%28x-13%29

Answer 11

well the center of the circle is(0,0) and the radius is 5 so the equation throught the point (13,00 is simply y=m9x-13) and where the m is the slope and the lenght of the perpendicular drawn from the centre on the line == and - 0-0=13m/suare root (1+m^2) and therefore m= +and - 5/12 and by substituting i got the answer