Question

How is the equation of this circle written in standard form?

x2 + y2 - 6x + 14y = 142

Answer 1

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

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=(√(200))^2\\\\(x-3)^2+(y+7)^2=(√(100\cdot2))^2\\\\(x-3)^2+(y+7)^2=(10\sqrt2)^2`

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