X2+5x+y2-y=-2
X2+2*5x2+(5/2)^2-(5/2)^2+y2-2*y/2+(1/2)^2-(1/2)^2=-2
(x+5/2)^2+(y-1/2)^2-13/2=-2
(x+5/2)^2+(y-1/2)^2=9/2
So centre =(-5/2,1/2)
Radius=(9/2)^(1/2)

7 0

3 years ago

Write the standard form of the equation of the line through the given point with the given slope. Through (-2,5), slope =-4

AleksandrR [38]

Answer:

$\huge\boxed{4x+y=-3}$

Step-by-step explanation:

The point-slope form of an equation of a line:

$y-y_1=m(x-x_1)$

(x₁, y₁) - point on a line

m - slope

We have

$m=-4,\ (-2,\ 5)\to x_1=-2,\ y_1=5$

Substitute:

$y-5=-4(x-(-2))\\\\y-5=-4(x+2)$

Convert to the standard form

$Ax+By=C$

$y-5=-4(x+2)$ use the distributive property

$y-5=(-4)(x)+(-4)(2)$

$y-5=-4x-8$ add 5 to both sides

$y-5+5=-4x-8+5$

$y=-4x-3$ add 4x to both sides

$4x+y=-4x+4x-3\\\\4x+y=-3$

3 0

3 years ago