My Math Forum finding a point on a circle given the slope of the tan line

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 September 30th, 2012, 03:22 PM #1 Senior Member   Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 finding a point on a circle given the slope of the tan line x^(2) + y^(2) = 100 mT = -3/4 Can someone show me how to go about finding this?
 September 30th, 2012, 03:48 PM #2 Senior Member   Joined: Jul 2011 Posts: 118 Thanks: 0 Re: finding a point on a circle given the slope of the tan l You can transform circle equation into function for upper and lower semicircle: $f_1=\sqrt{100-x^2}\\ f_2=-\sqrt{100-x^2}$ and now take derivative
 September 30th, 2012, 04:02 PM #3 Senior Member   Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 Re: finding a point on a circle given the slope of the tan l forgive me but wouldn't I take the Derv to find the slope of the tan line? I am trying to find a point on the upper half of the circle. I have tried to think on my own hence: rise/run and point slope y-y1=m(x-x1) when I to that I get (3/4, 0) would this be right?
September 30th, 2012, 04:43 PM   #4
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Re: finding a point on a circle given the slope of the tan l

Hello, mathkid!

Quote:
 $\text{Find the point on the upper half of the circle: }\:x^2\,+\,y^2 \:=\: 100 \;\;\;\text{where the slope of the tangent line is }-\frac{3}{4}$

$\text{W\!e have the equation: }\:y \:=\:(100\.-\,x^2)^{\frac{1}{2}}$

$\text{The derivative (slope) is: }\:y' \:=\:\frac{1}{2}(100\,-\,x^2)^{-\frac{1}{2}}(-2x) \;=\;\frac{-x}{\sqrt{100\,-\,x^2}}$

$\text{So we have: }\:\frac{-x}{\sqrt{100\,-\,x^2}} \:=\:-\frac{3}{4} \;\;\;\Rightarrow\;\;\;4x \:=\:3\sqrt{100\,-\,x^2}$

$\text{Square: }\:16x^2 \:=\:9(100\,-\,x^2) \;\;\;\Rightarrow\;\;\;16x^2 \:=\:900\,-\,9x^2 \;\;\;\Rightarrow\;\;\;25x^2 \:=\:900$

$\;\;\;x^2 \:=\:36 \;\;\;\Rightarrow\;\;\;x \:=\:\pm6$

$\text{If }x \,=\,-6,\:y#39; \,=\,+\frac{3}{4}\;\text{(wrong sign)}$

$\text{Therefore: }\:x \,=\, 6,\:y \,=\, 8$

 September 30th, 2012, 05:01 PM #5 Senior Member   Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 Re: finding a point on a circle given the slope of the tan l Wow!
 September 30th, 2012, 08:46 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: finding a point on a circle given the slope of the tan l Another approach: $x^2+y^2=100$ Implicitly differentiate with respect to x: $2x+2y\frac{dy}{dx}=0$ $\frac{dy}{dx}=-\frac{x}{y}$ Equate this to $-\frac{3}{4}$ $y=\frac{4}{3}x$ Substitute into original equation: $x^2+$$\frac{4}{3}x$$^2=100$ $x^2=36$ Take positive root. $x=6$ $y=8$
 October 1st, 2012, 06:49 PM #7 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: finding a point on a circle given the slope of the tan l You don't have to differentiate at all. Any tangent line to a circle is perpendicular to the radius at that point. If the slope of the tangent line is -3/4, then the radius at that point has slope 4/3. But if (x,y) is a point on the circle, then the slope of the line from the center, the origin (0, 0), is y/x. That is, y/x= 4/3 and, of course, $x^2+ y^2= 1$. From y/x= 4/3, y= (4/3)x so $x^2+ y^2= x^2+ (16/9)x^2= (1+ 16/9)x^2= (25/9)x^2= 1$ or $x^2= 9/25$, $x= \pm 3/5$ and then $y= \pm 4/3$.

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