What is the equation of the circle with center (2, –5) that passes through the point (–2, 10)?
1 answer:
The standard equation of a circle is
(x-h)^2 + (y-k)^2 = r^2
where the center is at point (h,k)
From the statement of the problem, it is already established that h = 2 and k = -5. What we have to determine is the value of r. This could be calculated by calculating the distance between the center and point (-2,10). The formula would be
r = square root [(x1-x2)^2 + (y1-y2)^2)]
r = square root [(2--2)^2 + (-5-10)^2)]
r = square root (241)
r^2 = 241
Thus, the equation of the circle is
(x-h)^2 + (y-k)^2 = r^2
where the center is at point (h,k)
From the statement of the problem, it is already established that h = 2 and k = -5. What we have to determine is the value of r. This could be calculated by calculating the distance between the center and point (-2,10). The formula would be
r = square root [(x1-x2)^2 + (y1-y2)^2)]
r = square root [(2--2)^2 + (-5-10)^2)]
r = square root (241)
r^2 = 241
Thus, the equation of the circle is
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