[Math] Find equation of tangent to the circle

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Closed 5 years ago.

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Coming back to this after about 6 months I now know how to solve it.

First I found the gradient of the radius $\frac{changeiny}{changeinx}$ >> $\frac{-8}{6}$ >> $1.33333333$

Then I found the negative reciprocal of the radius gradient to get the gradient of the tangent because it is at $90^\circ$ to the radius which becomes $\frac{1}{1.33333333}$ >> $0.75$

I then substituted the $x$ and $y$ values where the tangent touches the circle into the $y=mx+c$ equation >> $-8=0.75*6+c$ >> $-8=4.5+c$ >> $c=-12.5$

Which means that the equation of the tangent turned out to be ….. $y=0.75x-12.5$

ORIGINAL QUESTION:

The circle has the equation $x^2+y^2=100$

Find the equation of the tangent to the circle at the point $A(6,-8)$.

Any clues on how to solve this?

Answer 1

Since the shape is a circle, the tangent line at a point on the circle is perpendicular to the line connecting that point to the origin. The slope of the line from the origin to $A$ is clearly $\frac{-4}{3}$ since the "rise" from $(0,0)$ to $A=(6,-8)$ is $-8$ and the "run" is $6$. Thus, the slope of the perpendicular line is $\frac{3}{4}$. So the slope of your tangent line is $\frac{3}{4}$, and a point on the line is $A=(6,-8)$.