Answers:
7) Center= (-1,2) Radius= [tex]\boldsymbol{\sqrt{8}}[/tex] Equation: [tex](x+1)^2+(y-2)^2 = 8[/tex]
8) Center= (3,13) Radius= 13 Equation: [tex](x-3)^2+(y-13)^2 = 169[/tex]
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Explanation:
Problem 7
Let's find the distance from (-1,2) to (-3,4)
[tex](x_1,y_1) = (-1,2) \text{ and } (x_2, y_2) = (-3,4)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-1-(-3))^2 + (2-4)^2}\\\\d = \sqrt{(-1+3)^2 + (2-4)^2}\\\\d = \sqrt{(2)^2 + (-2)^2}\\\\d = \sqrt{4 + 4}\\\\d = \sqrt{8}\\\\[/tex]
This is the radius because it stretches from the center to a point on the circle, so [tex]r = \sqrt{8}[/tex]
Squaring both sides will get us [tex]r^2 = 8[/tex]
One useful template for a circle is the equation [tex](x-h)^2+(y-k)^2 = r^2\\\\[/tex]
(h,k) is the center
r is the radius
Let's plug in the given center (h,k) = (-1,2) and the r^2 value we found earlier.
[tex](x-h)^2+(y-k)^2 = r^2\\\\(x-(-1))^2+(y-2)^2 = 8\\\\(x+1)^2+(y-2)^2 = 8\\\\[/tex]
You can confirm this by using a tool like Desmos. See below.
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Problem 8
The endpoints of the diameter are (-2,1) and (8,25)
The center is the midpoint of these endpoints.
The midpoint of the x coordinates is (-2+8)/2 = 3
The midpoint of the y coordinates is (1+25)/2 = 13
The center is (h,k) = (3,13)
Now find the distance from the center to one of the points on the circle, let's say to (8,25)
[tex](x_1,y_1) = (3,13) \text{ and } (x_2, y_2) = (8,25)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(3-8)^2 + (13-25)^2}\\\\d = \sqrt{(-5)^2 + (-12)^2}\\\\d = \sqrt{25 + 144}\\\\d = \sqrt{169}\\\\d = 13\\\\[/tex]
The radius is exactly 13 units.
So,
[tex](x-h)^2+(y-k)^2 = r^2\\\\(x-3)^2+(y-13)^2 = 13^2\\\\(x-3)^2+(y-13)^2 = 169\\\\[/tex]
is the equation of this particular circle.
Visual confirmation is shown below.
