Question

Write the equation of the circle by completing the square. x'2 + 4x + y'2 - 2y = 20

Answer 1

$\bf \textit{let's start by grouping} \\\\\\ x^2+4x+y^2-2y=20\implies (x^2+4x)+(y^2-2y)=20 \\\\\\ (x^2+4x+\boxed{?}^2)+(y^2-2y+\boxed{?}^2)=20$

so hmm we're missing a couple of folks, one in each group, to get a perfect square trinomial

now, recall $\bf \begin{array}{cccccllllll} {{ a}}^2& + &2{{ a}}{{ b}}&+&{{ b}}^2\\ \downarrow && &&\downarrow \\ {{ a}}&& &&{{ b}}\\ &\to &({{ a}} + {{ b}})^2&\leftarrow \end{array}\qquad % perfect square trinomial, negative middle term \begin{array}{cccccllllll} {{ a}}^2& - &2{{ a}}{{ b}}&+&{{ b}}^2\\ \downarrow && &&\downarrow \\ {{ a}}&& &&{{ b}}\\ &\to &({{ a}} - {{ b}})^2&\leftarrow \end{array}$

so.. the middle term of a perfect square trinomial, is really 2 * the other 2 guys

so hmm we can say 2 * x * ? = 4x, thus 2x * ? = 4x, thus cross-multiplying
? = 2

now, for the other, 2 * y * ? = 2y, thus 2y* ? = 2y thus ? = 1

so our guys are 2 and 1

now, bear in mind, all we're doing, is borrowing from our very good friend Mr Zero, 0, so, if we add whatever, we also have to subtract whatever

thus   $\bf (x^2+4x+2^2)+(y^2-2y+1^2)-1^2-2^2=20 \\\\\\(x^2+4x+4)+(y^2-2y+1)-5=20\\\\\\ \boxed{(x+2)^2+(y-1)^2=25}\\\\ -----------------------------\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad center\ ({{ h}},{{ k}})\qquad radius={{ r}}\\\\ -----------------------------\\\\ (x-(-2))^2+(y-1)^2=5^2\qquad center\ (-2,1)\qquad radius=5$