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medal! write an equation of the circle with center (-2,3) and radius 9
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Remember that the standard form of the equation of a circle is denoted by \[(x-h)^2 + (y-k)^2 = r^2\] where (h,k) are the coordinates of the center and r is the radius
so is it setup (-2 - 3)^2 + (-2 - 3)^2
Since you have the coordinates of the center and the radius then you'll have it as: \[(x-(-2))^2 +(y-3)^2 = 9^2\] Simplify
x2+4x+y2-6y-68
this that correct
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is ths*
Simplifying you'll have: \[(x+2)^2 +(y-3)^2 = 9^2\] \[x^2 + 4x + 4 + y^2 -6y + 9 = 81\] Simplify that further
you get x2+4x+y2-6y-68
Your answer should be an equation. It should be \[x^2 + 4x + y^2 - 6y - 68 = 0\]
Thank you! much appreciated
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