What is the standard form of the equation of a circle with its center at (2, -3) and passing through the point (-2, 0)? (x − 2)squared + (y + 3)squared = 5 (x + 2)squared + (y − 3)squared = 25 (x − 2)squared + (y + 3)squared = 25 (x − 2)squared − (y + 3)squared = 5

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What is the standard form of the equation of a circle with its center at (2, -3) and passing through the point (-2, 0)? (x − 2)squared + (y + 3)squared = 5 (x + 2)squared + (y − 3)squared = 25 (x − 2)squared + (y + 3)squared = 25 (x − 2)squared − (y + 3)squared = 5

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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the general form is (x - a)^" + (y - b)^2 = r^2 where (a,b) is the center and r is the radius
yes
so you should be able to work out the left side now by plugging in a = 2 and b = -3
and the radius will be the distance between the center and the point (-2,0)
thanks
yw - if you need more help let me know
  • rvc
:) all the best
In general, the standard form of the equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\). In this case, (h,k) = (2,3) and (x,y) = (-2,0). Plug both points in to the equation, then simplify and solve for r. Next re-write the standard form of the equation, except this time, insert the given center point and the value of r.
ok ill try, thanks

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