A circle of centre (5, - 4) passes through the point (5, 0) find the equation.
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A circle of centre  (5, -4) passes through the point  (5, 0) find the equation.

A. x2 + y2 + 10x + 8y + 25 = 0

B. x2 + y2 + 10x - 8y - 25 = 0

C. x2 + y2 - 10x + 8y + 25 = 0

D. x2 + y2 - 10x - 8y - 25 = 0

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Nigeria
15 September 2020

School not set Nigeria
16 September 2020
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The general equation for a circle is given by: (x - h)2 + (y - k)2 = r-------------------- eqn(1)

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

at center  (5, - 4), h = 5 and k = -4

at point (5, 0), x = 5 and y = 0

from eqn(1),

(5 - 5)+ (0 - (-4))2 = r2

42 = r2

Take the square root of both sides,

r = 4

To get the equation of the circle, we substitute h = 5, k = -4 and r = 4 into eqn(1)

(x - 5)2 + (y - (-4))2 = 42

(x - 5)2 + (y + 4)2 = 42

x2 - 10x + 25 + y2 + 8y + 16 = 16

x2 + y2 - 10x + 8y 25 = 0

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