Question

координаты центра к радиусу хор 9х2+9у2+42х-54у-95=0

Answer 1

$9x^2+9y^2+42x-54y-95=0\(9x^2+42x)+(9y^2-54y)=95\(3x+7)^2-49+(3y-9)^2-81=95\(3x+7)^2+(3y-9)^2=225\(3x+7)^2+(3y-9)^2=15^2\9(x+frac{7}{3})^2+9(y-3)^2=15^2\(x+frac{7}{3})^2+(y-3)^2=5^2\R=5,; centr ; ; okryznosti; (-frac{7}{3},3)$