Circles and parabolas assigned by conditions

Text

Carry out the following:

• (a) write the equation of the circle with centre $A(2;-1)$ and radius $2$;
• (b) find the tangents to that circle through $P(3;-3)$ (distance method);
• (c) write the equation of the parabola with vertex $V(2;2)$ through the origin;
• (d) write the equation of the parabola through $A(2;1)$ meeting the $x$-axis at $x=-1$ and $x=3$;
• (e) find centre and radius of the circle $x^2+y^2+3x-2y-8=0$.

Solution

(a) $(x-2)^2+(y+1)^2=4$.

(b) $y+3=m(x-3)$; $\dfrac{|-m-2|}{\sqrt{m^2+1}}=2\Rightarrow 3m^2-4m=0\Rightarrow m=0\ \vee\ m=\tfrac43$. Tangents $y=-3$ and $y=\tfrac43 x-7$.

(c) $y=a(x-2)^2+2$; through the origin $a=-\tfrac12$: $y=-\tfrac12 x^2+2x$.

(d) $y=a(x+1)(x-3)$; through $A(2;1)$: $a=-\tfrac13$: $y=-\tfrac13(x+1)(x-3)$.

(e) Centre $\left(-\tfrac32;1\right)$, $R=\sqrt{\tfrac94+1+8}=\dfrac{3\sqrt5}{2}\approx 3{,}35$.

$\boxed{a_{(c)}=-\tfrac12,\ a_{(d)}=-\tfrac13,\ C\left(-\tfrac32;1\right),\ R=\tfrac{3\sqrt5}{2}}$