0

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

A. x^{2 }+ y^{2} + 10x + 8y + 25 = 0

B. x^{2 }+ y^{2 }+ 10x - 8y - 25 = 0

C. x^{2 }+ y^{2 }- 10x + 8y + 25 = 0

D. x^{2 }+ y^{2 }- 10x - 8y - 25 = 0

**Comments**

0

The general equation for a circle is given by: (x - h)^{2} + (y - k)^{2} = r^{2 }-------------------- 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)^{2 }+ (0 - (-4))^{2} = r^{2}

4^{2} = r^{2}

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} = 4^{2}

(x - 5)^{2} + (y + 4)^{2} = 4^{2}

x^{2} - 10x + 25 + y^{2} + 8y + 16 = 16

x^{2} + y^{2 }- 10x + 8y 25 = 0

**Answer: C**

Your Answer