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Determine the equation of the circle with center (0,−1) containing the point ( − 8 , 10 )

User Markijbema
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To find the equation of the circle with center (0, −1) that contains the point (−8, 10), we can use the standard form of the equation of a circle:


(x - h)^2 + (y - k)^2 = r^2

where (h, k) is the center of the circle and r is the radius.



Since the center of the circle is given as (0, −1), we have:


(x - 0)^2 + (y + 1)^2 = r^2


We still need to find the radius, r. We know that the circle passes through the point (−8, 10), so we can substitute these values for x and y and solve for r:


(-8 - 0)^2 + (10 + 1)^2 = r^2

64 + 121 = r^2

r^2 = 185

So the equation of the circle is:

x^2 + (y + 1)^2 = 185



Alternatively, we can expand the equation to get it in standard form:


x^2 + y^2 + 2y + 1 = 185

x^2 + y^2 + 2y - 184 = 0


So another form of the equation is:


x^2 + (y + 1)^2 =
185

User Marc B
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