Respuesta :
well, we know the center of the circle is at (-7, -1), and we also know a point on it is (8, 7), hmmm what's the distance between those points anyway?
well, a distance from a point on the circle and the center is namely the definition of its radius.
[tex] \bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\(\stackrel{x_1}{-7}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{7})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\\stackrel{radius}{r}=\sqrt{[8-(-7)]^2+[7-(-1)]^2}\implies r=\sqrt{(8+7)^2+(7+1)^2}\\\\\\r=\sqrt{15^2+8^2}\implies r=\sqrt{289}\implies r=17 [/tex]
so, we know the distance from a point to the center.... ok, so the point (-15, y) is on the circle too, therefore it must have the same distance "r" too,
[tex] \bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\\stackrel{center}{(\stackrel{x_1}{-7}~,~\stackrel{y_1}{-1})}\qquad (\stackrel{x_2}{-15}~,~\stackrel{y_2}{y})\\\\\\\stackrel{r}{17}=\sqrt{[-15-(-7)]^2+[y-(-1)]^2}\\\\\\17=\sqrt{(-15+7)^2+(y+1)^2}\implies 17^2=(-15+7)^2+(y+1)^2\\\\\\289=(-8)^2+(y+1)^2\implies 289=64+(y+1)^2\\\\\\225=(y+1)^2\implies \sqrt{225}=y+1\implies 15=y+1\implies 14=y [/tex]
Answer:
[tex] (x+7)^ +(y+1)^2 = r^2[/tex]
We know a point given who lies on the circel (x =8, y=7) and if we replace we got:
[tex] (8+7)^2 + (7+1)^2 = r^2[/tex]
[tex] 289 = r^2[/tex]
And if we take the square root we got:
[tex] r = \sqrt{289}= 17[/tex]
[tex] (-15+7)^2 +(y+1)^2 = 17^2[/tex]
And if we simplify we got:
[tex] 64 +(y+1)^2 = 289[/tex]
We can subtract 64 on both sides and we got:
[tex] (y+1)^2 = 225[/tex]
And we can apply square roots on both side and we got:
[tex] y+1= \pm \sqrt{225} = \pm 15[/tex]
And solving for y we got:
[tex] y = 15-1 = 14[/tex]
[tex] y = -15-1 = -16[/tex]
Step-by-step explanation:
For this case we can use the general formula for a circle given by:
[tex] (x-h)^2 +(y-k)^2 = r^2 [/tex]
Where [tex](h,k) [/tex] represent the center and r the radius. And for this special case we know this: [tex] (h = -7, k=-1)[/tex] and if we replace we got:
[tex] (x-(-7))^2 +(y-(-1))^2 = r^2 [/tex]
And if we simplify we got:
[tex] (x+7)^ +(y+1)^2 = r^2[/tex]
We know a point given who lies on the circel (x =8, y=7) and if we replace we got:
[tex] (8+7)^2 + (7+1)^2 = r^2[/tex]
[tex] 289 = r^2[/tex]
And if we take the square root we got:
[tex] r = \sqrt{289}= 17[/tex]
For the other part of the problem we know that x = -15 and we need to find the coordinate of y, for a point who lies on the circle, and we can do this:
[tex] (-15+7)^2 +(y+1)^2 = 17^2[/tex]
And if we simplify we got:
[tex] 64 +(y+1)^2 = 289[/tex]
We can subtract 64 on both sides and we got:
[tex] (y+1)^2 = 225[/tex]
And we can apply square roots on both side and we got:
[tex] y+1= \pm \sqrt{225} = \pm 15[/tex]
And solving for y we got:
[tex] y = 15-1 = 14[/tex]
[tex] y = -15-1 = -16[/tex]
