Answer:
Outside the circle
Step-by-step explanation:
Let's first write the equation of this circle: [tex](x-h)^2+(y-k)^2=r^2[/tex], where (h, k) is the center and r is the radius. Here, the center is (-6, -2). We need to find the radius, which will just be the distance from N to E:
NE = [tex]\sqrt{(-6-(-1))^2+(-2-1)^2} =\sqrt{25+9} =\sqrt{34}[/tex]
The radius is √34, which means that r² = 34. So, our equation is:
(x + 6)² + (y + 2)² = 34
Plug in -10 for x and -7 for y:
(x + 6)² + (y + 2)² = 34
x² + 12x + 36 + y² + 4y + 4 = 34
x² + 12x + y² + 4y + 40 = 34
x² + 12x + y² + 4y + 6 = 0
(-10)² + 12 * (-10) + (-7)² + 4 * (-7) + 6 = 7
Since 7 > 0, we know that H lies outside the circle.
