The diameter of a circle has endpoints P(–10, –2) and Q(4, 6).
a.Find the center of the circle.
b.Find the radius. If your answer is not an integer, express it in radical form.
c.Write an equation for the circle.
The centre must be the midpoint of PQ
which would be C( (-10+4)/2 , (-2+6)/2 )
= C(-3,2)
radius is √( (4+3)^2 + (2-6)^2) = √65
equation:
(x+3)^2 + (y-2)^2 = 65
@Connexus yep just goes to show how broken and lazy the educational system is. I see the irony in my sentence btw. We wouldn't cheat if they actually put effort into teaching new concepts instead of regurgitating the same things my grand pappy learned in the 60s.
P(-10, -2), C(x, y), Q(4, 6).
a. x+10 = 1/2(4+10)
X = -3.
y+2 = 1/2(6+2)
Y = 2.
b. Radius = sqrt ((-3+10)^2+(2+2)^2) = sqrt 65.
c. (x+10)^2+(y+2)^2 = 65.
omg this is from 2012
a. To find the center of the circle, we use the midpoint formula. ((-10+4)/2 , (-2+6)/2 )= C(-3,2). The center is at (-3,2).
b. d = √((x_2 - x_1)^2 + (y_2-y_1)^2)
d=√((-3 - 4)^2 + (2-6)^2)
d=√((-7)^2 + (-4)^2)
d=√(49 + 16)
d=√(65)
Radius=√(65)
c. (x+3)^2+(y-2)^2 = 65.
first one is right
1. 19.6
2. 64
3. 243
4. 58.5
5. 53.5
6. 73
7. 38
8. 54
9. x=15.8
10. 45.5
11. B (dot right below 4 on the x axis and a small circle)
12. 28
13. 11.6
14. 48
15. x=61
16. x=15 and none of the triangles are congruent, the triangle is scalene
17. source is (-6,-4) and the range is 6
18. C( (-10+4)/2 , (-2+6)/2 )= C(-3,2)