Asked by Regina
what is the equation of the circle with center (-6,7) that passes through the point (4,-2) ?
Answer choices:
(x+6)^2 + (y-2)^2=9
(x-3)^2 + (y+2)^2=9
(x-3)^2+(y+2)^2=3
(x=3)^2=(y-2)^2=3
Answer choices:
(x+6)^2 + (y-2)^2=9
(x-3)^2 + (y+2)^2=9
(x-3)^2+(y+2)^2=3
(x=3)^2=(y-2)^2=3
Answers
Answered by
Steve
clearly, the circle is
(x+6)^2 + (y-7)^2 = r^2
So, I suspect a typo. Check your coordinates again, and recall that a circle with center at (h,k) is
(x-h)^2 + (y-k)^2 = r^2
Once you fix your error, just find how far away it is from (4,-2).
(x+6)^2 + (y-7)^2 = r^2
So, I suspect a typo. Check your coordinates again, and recall that a circle with center at (h,k) is
(x-h)^2 + (y-k)^2 = r^2
Once you fix your error, just find how far away it is from (4,-2).
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