How is the equation of this circle written in standard form?x2 + y2 - 6x + 14y = 142
Accepted Solution
A:
Answer:[tex]\large\boxed{(x-3)^2+(y+7)^2=200\to(x-3)^2+(y+7)^2=(10\sqrt2)^2}[/tex]Step-by-step explanation:The standard form of an equation of a circle:[tex](x-h)^2+(y-k)^2=r^2[/tex](h, k) - centerr - radiusWe have the equation:[tex]x^2+y^2-6x+14y=142[/tex]We must use[tex](a\pm b)^2=a^2\pm2ab+b^2[/tex][tex]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[/tex][tex]center:(3,\ -7)\\radius:10\sqrt2[/tex]