Tangents to a circle as the intercept varies

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Consider the line $y=x+q$ and the circle of centre $C(0;0)$ 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.
• (d) Draw the two tangents found.

Solution

(a) $x^2+y^2=9$.

(c) Distance method. The line is $x-y+q=0$; requiring the distance from the origin to equal $R=3$: $\frac{|q|}{\sqrt{1+1}}=3 \implies |q|=3\sqrt2.$ Check with $\Delta$: substituting $y=x+q$ into $x^2+y^2=9$ gives $2x^2+2qx+q^2-9=0$, and $\tfrac{\Delta}{4}=q^2-2(q^2-9)=18-q^2=0 \implies q^2=18$. With $3\sqrt2\approx4.243$: $\boxed{\ q=\pm3\sqrt2\approx\pm4.24\ }$