The standard equation of a circle with the center at (h, k) and a radius of r units is (x – h)²+ (y – k)² = r². Given the cross section which is a circle, find the equation of a circle in standard form with center at (0, -5) and one point on the circle is (2, 3).
show your solution

[tex]r = \sqrt{(x2 - x1)^{2} + (y2 - y1)^{2} } \\ r = \sqrt{ {(2 - 0)}^{2} + {(3 - ( - 5))}^{2} } \\ r = \sqrt{ {(2)}^{2} + {(3 + 5)}^{2} } \\ r = \sqrt{4 + 64} \\ r = \sqrt{68} [/tex]

(3) Substitute the coordinates of the center (0,-5) and the obtained r value in the standard form of the equation

The standard equation of a circle with the center at (h, k) and a radius of r units is (x – h)²+ (y – k)² = r². Given the cross section which is a circle, find the equation of a circle in standard form with center at (0, -5) and one point on the circle is (2, 3).
show your solution

Sagot :

x² + (y + 5)² = (√68)²

x1 = 0

y1 = -5

x2 = 2

y2 = 3

(x - h)² + (y - k)² = r²

(x - 0)² + (y -(-5)) = (√68)²

x² + (y + 5)² = (√68)²

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