Question

How is the equation of this circle written in standard form? x2 + y2 - 6x + 14y = 142

Answer 1

Answer:

$\large\boxed{(x-3)^2+(y+7)^2=200\to(x-3)^2+(y+7)^2=(10\sqrt2)^2}$

Step-by-step explanation:

The standard form of an equation of a circle:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have the equation:

$x^2+y^2-6x+14y=142$

We must use

$(a\pm b)^2=a^2\pm2ab+b^2$

$x^2-6x+y^2+14y=142\\\\x^2-2(x)(3)+y^2+2(y)(7)=142\qquad\text{add}\ 3^2\ \text{and}\ 7^2\ \text{to both sides}\\\\\underbrace{x^2-2(x)(3)+3^2}_{(a-b)^2=a^2-2ab+b^2}+\underbrace{y^2+2(y)(7)+7^2}_{(a+b)^2=a^2+2ab+b^2}=142+3^2+7^2\\\\(x-3)^2+(y+7)^2=142+9+49\\\\(x-3)^2+(y+7)^2=200\\\\(x-3)^2+(y+7)^2=(\sqrt{200})^2\\\\(x-3)^2+(y+7)^2=(\sqrt{100\cdot2})^2\\\\(x-3)^2+(y+7)^2=(10\sqrt2)^2$

$center:(3,\ -7)\\radius:10\sqrt2$