Additional Mathematics – Coordinate geometry of the circle | e-Consult

The circle has centre \(O(0,0)\) and radius \(r=5\).

If the line \(y=mx\) is tangent, the distance from the centre to the line must equal the radius.

Distance from \(O\) to \(y=mx\) (or \(mx-y=0\)) is

\[

\frac{|0|}{\sqrt{m^{2}+1}}=0\quad\text{(not useful)}.

\]

Instead, use the condition that the line meets the circle in exactly one point. Substitute \(y=mx\) into the circle:

\[

x^{2}+(mx)^{2}=25\;\Longrightarrow\;x^{2}(1+m^{2})=25.

\]

For a single intersection, the quadratic in \(x\) must have a double root, i.e. discriminant zero – but here it is already solved for \(x^{2}\). Hence we require \(x^{2}\) to be positive and unique, which occurs when

\[

x^{2}=\frac{25}{1+m^{2}}.

\]

The point of tangency is \(\bigl(x{0},y{0}\bigr)=\Bigl(\frac{5}{\sqrt{1+m^{2}}},\; \frac{5m}{\sqrt{1+m^{2}}}\Bigr).\)

The radius \(OC\) is perpendicular to the tangent. The slope of \(OC\) is \(\dfrac{y{0}}{x{0}}=m\). Therefore the tangent at \((x{0},y{0})\) has slope \(-\dfrac{1}{m}\) (provided \(m\neq0\)).

Using point‑slope form:

\[

y-y{0}=-\frac{1}{m}\,(x-x{0}).

\]

Substituting \(x{0},y{0}\) and simplifying gives the tangent equation

\[

\boxed{mx+y=5}.

\]

Thus the line \(y=mx\) is tangent to the circle exactly when it satisfies \(mx+y=5\). Solving together with \(y=mx\) yields \(m^{2}+1=5/m\) → \(m^{2}= \frac{5}{m}-1\). The only real solution is \(m=\pm\frac{3}{4}\). Hence the two possible tangents are

• \( \displaystyle \frac{3}{4}x+y=5\)
• \( \displaystyle -\frac{3}{4}x+y=5\)