A circle with centre (5,-4) passes through the point (5, 0). Find its 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
waec 2020furthermathematics

The correct answer is C. x\(^2\) + y\(^2\) - 10x + 8y + 25 =0

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

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

x\(^2\) - 2(3)x + y\(^2\) - 2(-4) y + 5\(^2\) + (-4)\(^2\) = r\(^2\)

x\(^2\) - 10x + y\(^2\) + 8y + 25 + 16 = r\(^2\)

x\(^2\) - 10x + y\(^2\) + 8y + 41 = r\(^2\)

at point (5,0)

5\(^2\) - 10(5) + 0\(^2\) + 8(0) + 41 = r\(^2\) 

25 - 50 + 41 = r\(^2\)

16 = r\(^2\) 

r = \(\sqrt{16}\) 

= 4 

x\(^2\) + y\(^2\) - 10x + 8y + 25 = 0

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