Circle A has a center at (1 ,1 ) and a radius of 1 . Circle B has a center at (2 ,-3 ) and a radius of 5 . Do the circles overlap? If not, what is the smallest distance between them?

Mar 2, 2016

points of intersection: $\left(\frac{27 + 4 \sqrt{19}}{34} , \frac{62 + \sqrt{19}}{34}\right) \mathmr{and} \left(\frac{27 - 4 \sqrt{19}}{34} , \frac{62 - \sqrt{19}}{34}\right)$ or approximately $\left(1.307 , 1.952\right)$ and $\left(0.281 , 1.695\right)$.

Explanation:

Recall that the general equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$

where:
$x =$any x-coordinate of the circle
$h =$x-coordinate of the circle's center
$y =$any y-coordinate of the circle
$k =$y-coordinate of the circle's center
$r =$radius

$1$. Determine the equation of each circle:

Circle A
${\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2} = {1}^{2}$

Circle B
${\left(x - 2\right)}^{2} + {\left(y + 3\right)}^{2} = {5}^{2}$

$2$. Simplify the equations so that they both equal $0$.

Circle A
${\left(x - 1\right)}^{2} + {\left(y - 1\right)}^{2} = {1}^{2}$

$\left(x - 1\right) \left(x - 1\right) + \left(y - 1\right) \left(y - 1\right) = 1$

$\left({x}^{2} - 2 x + 1\right) + \left({y}^{2} - 2 y + 1\right) = 1$

Equation $1$: $\textcolor{\mathmr{and} a n \ge}{{x}^{2} + {y}^{2} - 2 x - 2 y + 1 = 0}$

Circle B
${\left(x - 2\right)}^{2} + {\left(y + 3\right)}^{2} = {5}^{2}$

$\left(x - 2\right) \left(x - 2\right) + \left(y + 3\right) \left(y + 3\right) = 25$

$\left({x}^{2} - 4 x + 4\right) + \left({y}^{2} + 6 y + 9\right) = 25$

Equation $2$: $\textcolor{b l u e}{{x}^{2} + {y}^{2} - 4 x + 6 y - 12 = 0}$

$3$. Since the equations both equal $0$, they must both be equal to each other.

$\textcolor{\mathmr{and} a n \ge}{{x}^{2} + {y}^{2} - 2 x - 2 y + 1} = \textcolor{b l u e}{{x}^{2} + {y}^{2} - 4 x + 6 y - 12}$

$4$. Simplify.

$\textcolor{red}{\cancel{\textcolor{\mathmr{and} a n \ge}{{x}^{2}}}} \textcolor{red}{\cancel{\textcolor{\mathmr{and} a n \ge}{+ {y}^{2}}}} \textcolor{\mathmr{and} a n \ge}{- 2 x - 2 y + 1} = \textcolor{red}{\cancel{\textcolor{b l u e}{{x}^{2}}}} \textcolor{red}{\cancel{\textcolor{b l u e}{+ {y}^{2}}}} \textcolor{b l u e}{- 4 x + 6 y - 12}$

$\textcolor{\mathmr{and} a n \ge}{- 2 x - 2 y + 1} = \textcolor{b l u e}{- 4 x + 6 y - 12}$

$5$. Solve for $y$.

$\textcolor{\mathmr{and} a n \ge}{- 2 y}$ $\textcolor{b l u e}{- 6 y} = \textcolor{b l u e}{- 4 x}$ $\textcolor{\mathmr{and} a n \ge}{+ 2 x}$ $\textcolor{b l u e}{- 12}$ $\textcolor{\mathmr{and} a n \ge}{- 1}$

$- 8 y = - 2 x - 13$

$\frac{- 8 y}{- 8} = \frac{- 2 x}{- 8} - \frac{13}{- 8}$

$\textcolor{p u r p \le}{y = \frac{1}{4} x + \frac{13}{8}}$

$6$. Substitute $\textcolor{p u r p \le}{y = \frac{1}{4} x + \frac{13}{8}}$ into either equation $1$ or $2$. In this case, we'll use equation $1$.

$\textcolor{\mathmr{and} a n \ge}{{x}^{2} + {\textcolor{p u r p \le}{y}}^{2} - 2 x - 2 \textcolor{p u r p \le}{y} + 1 = 0}$

${x}^{2} + {\textcolor{p u r p \le}{\left(\frac{1}{4} x + \frac{13}{8}\right)}}^{2} - 2 x - 2 \textcolor{p u r p \le}{\left(\frac{1}{4} x + \frac{13}{8}\right)} + 1 = 0$

$7$. Simplify.

${x}^{2} + \textcolor{p u r p \le}{\left(\frac{1}{4} x + \frac{13}{8}\right) \left(\frac{1}{4} x + \frac{13}{8}\right)} - 2 x - 2 \textcolor{p u r p \le}{\left(\frac{1}{4} x + \frac{13}{8}\right)} + 1 = 0$

${x}^{2} + \left(\frac{1}{16} {x}^{2} + \frac{13}{16} x + \frac{169}{64}\right) - 2 x - {\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}^{1} \textcolor{p u r p \le}{\left(\frac{1}{\textcolor{red}{\cancel{\textcolor{p u r p \le}{4}}}} ^ 2 x + \frac{13}{\textcolor{red}{\cancel{\textcolor{p u r p \le}{8}}}} ^ 4\right)} + 1 = 0$

$\frac{17}{16} {x}^{2} + \frac{13}{16} x + \frac{169}{64} - 2 x - \frac{1}{2} x - \frac{13}{4} + 1 = 0$

$\frac{17}{16} {x}^{2} - \frac{27}{16} x + \frac{25}{64} = 0$

$8$. Use the quadratic formula to factor the equation. This will tell you the x-coordinates of where the two circles intersect.

$\textcolor{t u r q u o i s e}{a = \frac{17}{16}} \textcolor{w h i t e}{X X X} \textcolor{g o l d}{b = - \frac{27}{16}} \textcolor{w h i t e}{X X X} \textcolor{b r o w n}{c = \frac{25}{64}}$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- \left(\textcolor{g o l d}{- \frac{27}{16}}\right) \pm \sqrt{{\left(\textcolor{g o l d}{- \frac{27}{16}}\right)}^{2} - 4 \left(\textcolor{t u r q u o i s e}{\frac{17}{16}}\right) \left(\textcolor{b r o w n}{\frac{25}{64}}\right)}}{2 \left(\textcolor{t u r q u o i s e}{\frac{17}{16}}\right)}$

$x = \frac{\frac{27}{16} \pm \sqrt{\frac{729}{256} - \frac{425}{256}}}{\frac{17}{8}}$

$x = \frac{\frac{27}{16} \pm \sqrt{\frac{304}{256}}}{\frac{17}{8}}$

$x = \frac{\frac{27}{16} \pm \frac{\sqrt{19}}{4}}{\frac{17}{8}}$

$x = \left(\frac{27}{16} \pm \frac{4 \sqrt{19}}{16}\right) \times \frac{8}{17}$

$x = \left(\frac{27 \pm 4 \sqrt{19}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{16}}}} ^ 2\right) \times {\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}}^{1} / 17$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x = \frac{27 \pm 4 \sqrt{19}}{34} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$9$. Now that you have the x-coordinates of the points of intersection, substitute each x-coordinate into $\textcolor{p u r p \le}{y = \frac{1}{4} x + \frac{13}{8}}$ to find the corresponding y-coordinates.

Intersection 1
$\textcolor{p u r p \le}{y = \frac{1}{4} x + \frac{13}{8}}$

$y = \frac{1}{4} \left(\frac{27 + 4 \sqrt{19}}{34}\right) + \frac{13}{8}$

$y = \frac{27 + 4 \sqrt{19}}{136} + \frac{13}{8}$

$y = \frac{27 + 4 \sqrt{19}}{136} + \frac{17 \left(13\right)}{136}$

$y = \frac{27 + 4 \sqrt{19}}{136} + \frac{221}{136}$

$y = \frac{248 + 4 \sqrt{19}}{136}$

$y = \frac{4 \left(62 + \sqrt{19}\right)}{4 \left(34\right)}$

$y = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} \left(62 + \sqrt{19}\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} \left(34\right)}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} y = \frac{62 + \sqrt{19}}{34} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Intersection 2
$\textcolor{p u r p \le}{y = \frac{1}{4} x + \frac{13}{8}}$

$y = \frac{1}{4} \left(\frac{27 - 4 \sqrt{19}}{34}\right) + \frac{13}{8}$

$y = \frac{27 - 4 \sqrt{19}}{136} + \frac{13}{8}$

$y = \frac{27 - 4 \sqrt{19}}{136} + \frac{17 \left(13\right)}{136}$

$y = \frac{27 - 4 \sqrt{19}}{136} + \frac{221}{136}$

$y = \frac{248 - 4 \sqrt{19}}{136}$

$y = \frac{4 \left(62 - \sqrt{19}\right)}{4 \left(34\right)}$

$y = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} \left(62 - \sqrt{19}\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4}}} \left(34\right)}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} y = \frac{62 - \sqrt{19}}{34} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

If you graphed the two circles, it would look like:

Zoomed in, you can get a sense as to where the intersections are:

$\therefore$, the circles overlap at $\left(\frac{27 + 4 \sqrt{19}}{34} , \frac{62 + \sqrt{19}}{34}\right) \mathmr{and} \left(\frac{27 - 4 \sqrt{19}}{34} , \frac{62 - \sqrt{19}}{34}\right)$ or at approximately $\left(1.307 , 1.952\right)$ and $\left(0.281 , 1.695\right)$.