Asked by Tim
1. What is the equation of the circle with center (-4,-3) that passes through the point (6,2)?
A) (x-4)^2+(y-3)^2=25
B) (x-4)^2+(y-3)^2=125
C) (x-(-4))^2+(y-(-3))^2=25
D) (x-(-4))^2+(y-(-3))^2=25
I'm really not sure how to do this at all, so any help would be absolutely fantastic. Thank you so much.
When I solve the first two, I both get positive (4,3). Which isn't correct. And i'm confused.
A) (x-4)^2+(y-3)^2=25
B) (x-4)^2+(y-3)^2=125
C) (x-(-4))^2+(y-(-3))^2=25
D) (x-(-4))^2+(y-(-3))^2=25
I'm really not sure how to do this at all, so any help would be absolutely fantastic. Thank you so much.
When I solve the first two, I both get positive (4,3). Which isn't correct. And i'm confused.
Answers
Answered by
Reiny
in general if the centre is (h,k), the equation is
(x-h)^2 + (y-k)^2 = r^2
notice that the sign inside the bracket is opposite to the sign used in the given point.
so for (-4,-3) it must be
(x-(-4) )^2 + (y - (-3) )^2 = r^2
(x+4)^2 + (y+3)^2 = r^2
to find the r^2, just plug in the given point (6,2)
(6+4)^2 + (2+3)^2 = r^2
125 = r^2
by the looks of the choices , none of them are correct, unless you have a typo in either C or D
(notice they are the same)
Furthermore, "they" should not have left it as
(x-(-4))^2 etc
correct answer:
(x+4)^2 + (y+3)^2 = 125
(x-h)^2 + (y-k)^2 = r^2
notice that the sign inside the bracket is opposite to the sign used in the given point.
so for (-4,-3) it must be
(x-(-4) )^2 + (y - (-3) )^2 = r^2
(x+4)^2 + (y+3)^2 = r^2
to find the r^2, just plug in the given point (6,2)
(6+4)^2 + (2+3)^2 = r^2
125 = r^2
by the looks of the choices , none of them are correct, unless you have a typo in either C or D
(notice they are the same)
Furthermore, "they" should not have left it as
(x-(-4))^2 etc
correct answer:
(x+4)^2 + (y+3)^2 = 125
Answered by
Tim
Oh my apologies! D was 125, not 25. Thank you so much though, I was quiet confused as well since I was thinking they were both 25.
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