Tangent line to a circle with both methods

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Consider the line $y=\dfrac{x}{2}+q$ and the circle of centre $C(0;3)$ and radius $3$.

• (a) Find the equation of the circle and draw it.
• (b) Draw the line for $q=0$.
• (c) Find the two values of $q$ for which the line is tangent to the circle, using the $\Delta$ method.
• (d) Find the two values of $q$ again, using the distance method.
• (e) Draw the two tangents found.

Solution

(a) $x^2+(y-3)^2=9 \implies x^2+y^2-6y=0$.

(c) $\Delta$ method. Substituting $y=\tfrac{x}{2}+q$: $x^2+\Bigl(\tfrac{x}{2}+q-3\Bigr)^2=9 \implies \tfrac54 x^2+(q-3)x+(q-3)^2-9=0.$ Tangency requires $\Delta=0$: $(q-3)^2-4\cdot\tfrac54\bigl[(q-3)^2-9\bigr]=-4(q-3)^2+45=0 \implies (q-3)^2=\tfrac{45}{4}.$ $q=3\pm\frac{3\sqrt5}{2}.$

(d) Distance method. The line is $x-2y+2q=0$; the distance from the centre $C(0;3)$ must equal $R=3$: $\frac{|0-6+2q|}{\sqrt{1+4}}=3 \implies |2q-6|=3\sqrt5 \implies q=3\pm\frac{3\sqrt5}{2}.$ Both methods give the same result. With $\tfrac{3\sqrt5}{2}\approx3.354$: $\boxed{\ q_1=3+\tfrac{3\sqrt5}{2}\approx6.35, \qquad q_2=3-\tfrac{3\sqrt5}{2}\approx-0.35\ }$