How to Find a Point on a Tangent Line to a Circle

Key concept: A tangent line to a circle is perpendicular to the radius drawn to the point of tangency. That one fact turns a hard-looking coordinate geometry question into a short slope problem.

Step by step

Let the center of the circle be $(-1,1)$ and the point of tangency be $(5,-4)$.

1) Find the slope of the radius

The radius goes from $(-1,1)$ to $(5,-4)$:

$m_{radius}=\frac{-4-1}{5-(-1)}=\frac{-5}{6}$

2) Find the tangent slope

A tangent is perpendicular to the radius, so its slope is the negative reciprocal:

$m_{tangent}=\frac{6}{5}$

3) Write the line through $(5,-4)$

Using point-slope form:

$y+4=\frac{6}{5}(x-5)$

Simplify:

$y+4=\frac{6}{5}x-6$

$y=\frac{6}{5}x-10$

4) Test the choices

• A: $(0,\tfrac{6}{5})$ gives $y=-10$ when $x=0$, so no.
• B: $(4,7)$ gives $y=\tfrac{24}{5}-10=-\tfrac{26}{5}$, so no.
• C: $(10,2)$ gives $y=12-10=2$, so yes.
• D: $(11,1)$ gives $y=\tfrac{66}{5}-10=\tfrac{16}{5}$, so no.

Correct answer: C. $(10,2)$

Pitfall alert

A very common mistake is to use the slope of the radius as the tangent slope. They are not the same; they are negative reciprocals. Another issue is plugging the tangent point into the circle equation instead of the line equation—this question is asking for a point on the line, not on the circle.

Try different conditions

If the point of tangency were different, the tangent slope would still be the negative reciprocal of the radius slope. If the center changed, you would repeat the same two-step process: find the radius slope first, then write the tangent line through the tangency point and test the options.

Further reading

tangent line, perpendicular slopes, point-slope form