Tangent with Given Slope

Tangents to x² + y² = a² with slope m. Two tangents exist for every finite m; they are parallel and symmetric about the centre.

For the circle $x^2 + y^2 = a^2$, find all tangent lines with slope $m$.

Any line with slope $m$ has the form $y = mx + c$. From the tangent condition:

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

Therefore the two tangent lines are:

$$y = mx + a\sqrt{1 + m^2} \quad \text{and} \quad y = mx - a\sqrt{1 + m^2}$$

Combined: $y = mx \pm a\sqrt{1 + m^2}$.

Geometric interpretation: These two lines are parallel (same slope $m$), on opposite sides of the centre. The perpendicular distance from the origin to each is exactly $a$, confirming tangency.

Point of contact: Substituting $y = mx + a\sqrt{1+m^2}$ into the circle equation and solving gives the single contact point:

$$\left(\frac{-am}{\sqrt{1+m^2}},\, \frac{a}{\sqrt{1+m^2}}\right)$$

For the $-$ case, the contact point is the reflection of this in the origin.

Note on vertical tangents: The formula breaks down for a vertical line ($m$ undefined). The vertical tangents to $x^2 + y^2 = a^2$ are $x = \pm a$, obtained directly.