Condition for a Line to be Tangent

Line y = mx + c is tangent to x² + y² = a² iff c² = a²(1 + m²), equivalently the perpendicular distance from centre equals radius.

Consider the circle $x^2 + y^2 = a^2$ (centre $O$ at origin, radius $a$) and line $y = mx + c$.

A line is tangent to a circle if and only if it meets the circle at exactly one point, which is equivalent to the perpendicular distance from the centre to the line equalling the radius.

Method 1 — Distance condition:

Rewrite the line as $mx - y + c = 0$. Distance from $O(0, 0)$:

$$d = \frac{|c|}{\sqrt{m^2 + 1}}$$

Setting $d = a$:

$$\frac{|c|}{\sqrt{m^2 + 1}} = a \implies c^2 = a^2(1 + m^2)$$

Method 2 — Discriminant condition:

Substitute $y = mx + c$ into $x^2 + y^2 = a^2$:

$$x^2 + (mx + c)^2 = a^2$$ $$(1 + m^2)x^2 + 2mcx + c^2 - a^2 = 0$$

For exactly one solution, discriminant $= 0$:

$$4m^2c^2 - 4(1 + m^2)(c^2 - a^2) = 0$$ $$m^2c^2 - c^2 + a^2 - m^2c^2 + m^2a^2 = 0$$ $$c^2 = a^2(1 + m^2)$$

Both methods give the same result. The distance method is faster; the discriminant method is more algebraically transparent.