Circle, intersections with a line and parallel tangents

Text

Consider the circle with centre $C(3;2)$ and radius $5$.

• (a) draw it;
• (b) find its equation;
• (c) find the intersection between the circle and the line $y=\dfrac{1}{2}x+3$;
• (d) find the equations of the two tangent lines to the circle that are parallel to $y=\dfrac{1}{2}x+3$.

Solution

(a)–(b) From $(x-x_C)^2+(y-y_C)^2=r^2$: $(x-3)^2+(y-2)^2=25\ \Longleftrightarrow\ x^2+y^2-6x-4y-12=0.$

(c) Substituting $y=\tfrac12 x+3$ into the equation: $x^2+\left(\tfrac12 x+3\right)^2-6x-4\left(\tfrac12 x+3\right)-12=0\Rightarrow \tfrac54 x^2-5x-15=0\Rightarrow x^2-4x-12=0.$ Hence $x=6$ or $x=-2$, giving the points $A(6;6)$ and $B(-2;2)$.

(d) The parallel tangents have slope $\tfrac12$: $y=\tfrac12 x+q$, i.e. $x-2y+2q=0$. Imposing distance from $C$ equal to $5$: $\frac{|3-2\cdot 2+2q|}{\sqrt{1^2+2^2}}=\frac{|2q-1|}{\sqrt5}=5\Rightarrow |2q-1|=5\sqrt5\Rightarrow q=\frac{1\pm5\sqrt5}{2}.$ Tangents $y=\tfrac12 x+\dfrac{1+5\sqrt5}{2}\approx \tfrac12 x+6.09$ and $y=\tfrac12 x+\dfrac{1-5\sqrt5}{2}\approx \tfrac12 x-5.09$.

$\boxed{x^2+y^2-6x-4y-12=0;\ A(6;6),\ B(-2;2);\ q=\tfrac{1\pm5\sqrt5}{2}}$