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What is an equation of the tangent line to the circle $(x-3)^2+(y-5)^2=5$ at $P(4,7)$ ?

What I've found:

  1. Center of the circle : $(3,5)$

  2. Radius : $\sqrt5$

The equation will be in $y=mx+b$ format, therefore I need to find the slope and $y$-intercept, is it correct? How can I find the slope and the intercept?

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  • $\begingroup$ I think I found the answer. y=-1/2x+13/2 is it right? $\endgroup$
    – han zo
    Aug 26, 2021 at 0:50
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    $\begingroup$ please edit the question to show your work / steps so we can check. $\endgroup$
    – Math Lover
    Aug 26, 2021 at 0:52
  • $\begingroup$ @hanzo The slope (gradient) is right but the intercept is not. You can see your solution is not correct by putting in $x=4$. You would expect $y=7$ but you don't get that. Show your detailed working. $\endgroup$
    – Deepak
    Aug 26, 2021 at 0:55
  • $\begingroup$ $(x+4-3)^2+(y+7-5)^2=5$ reduces to $x^2+2x+y^2+4x=0$ i.e. the tangent line is $2(x-4)+4(y-7)=0.$ $\endgroup$ Aug 26, 2021 at 0:55
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    $\begingroup$ @peterwhy I see. Does it mean that I should put P(4,7)? Then it gives an equation of y=-1/2x+9, right? $\endgroup$
    – han zo
    Aug 26, 2021 at 1:03

1 Answer 1

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The centre C is (3,5). The tangent line is perpendicular to the line passing through the centre and the point of tangency. Hence the slope of tangent will be the negative reciprocal of slope of CP. The tangent passes through point P so you get the value of it's y-intercept.

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  • $\begingroup$ If you think a post answers your question, you can click on the tick sign just below the downvote sign ....it gives you credit. $\endgroup$
    – Ilovemath
    Aug 29, 2021 at 9:23

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