Math • April 2026

Finding the Tangency Condition of
a Line to a Circle

A
Written By Archive Editorial
Reading Time 5 Min Read

Understanding Tangency

A line is considered tangent to a circle if it touches the circle at exactly one point. Geometrically, this means that the shortest perpendicular distance from the center of the circle to the line must be exactly equal to the radius of the circle.

Problem Statement

We want to find the condition under which the line $px + qy = r$ is tangent to the circle $x^2 + y^2 = a^2$.

Step-by-Step Derivation

1. Identify Circle Properties

For the circle $x^2 + y^2 = a^2$:

  • The center is at the origin $(h, k) = (0, 0)$.
  • The radius is $r_{circle} = a$.

2. Identify Line Properties

The line equation is $px + qy - r = 0$. Comparing this to the standard form $Ax + By + C = 0$, we have:

  • $A = p$
  • $B = q$
  • $C = -r$

3. Apply the Perpendicular Distance Formula

The formula for the perpendicular distance ($d$) from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is: $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Substituting our values (point $(0,0)$, line $px + qy - r = 0$): $d = \frac{|p(0) + q(0) - r|}{\sqrt{p^2 + q^2}} = \frac{|-r|}{\sqrt{p^2 + q^2}} = \frac{|r|}{\sqrt{p^2 + q^2}}$

4. Set the Tangency Condition

For the line to be tangent, the perpendicular distance $d$ must equal the radius $a$: $\frac{|r|}{\sqrt{p^2 + q^2}} = a$

5. Final Simplification

Square both sides to remove the absolute value and the radical: $\frac{r^2}{p^2 + q^2} = a^2$ $r^2 = a^2(p^2 + q^2)$

Conclusion

The required condition for the line $px + qy = r$ to be tangent to the circle $x^2 + y^2 = a^2$ is $r^2 = a^2(p^2 + q^2)$.

Editorial Note

Automated explainer generated from student question.

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