# Find an equation of the tangent line to the circle $(x-3)^2+(y-5)^2=5$ at $P(4,7).$ [closed]

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?

• I think I found the answer. y=-1/2x+13/2 is it right? Aug 26, 2021 at 0:50
• please edit the question to show your work / steps so we can check. Aug 26, 2021 at 0:52
• @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. Aug 26, 2021 at 0:55
• $(x+4-3)^2+(y+7-5)^2=5$ reduces to $x^2+2x+y^2+4x=0$ i.e. the tangent line is $2(x-4)+4(y-7)=0.$ Aug 26, 2021 at 0:55
• @peterwhy I see. Does it mean that I should put P(4,7)? Then it gives an equation of y=-1/2x+9, right? Aug 26, 2021 at 1:03