Find the radius of the circle (x^{2} + y^{2} – 8x – 2y + 1 = 0).
-
A.
9 -
B.
7 -
C.
4 -
D.
3
Correct Answer: Option C
Explanation
Given the equation of the circle (x^{2} + y^{2} – 8x – 2y + 1 = 0).
The equation of a circle is given as ((x – a)^{2} + (y – b)^{2} = r^{2})
Expanding, we have (x^{2} – 2ax + a^{2} + y^{2} – 2by + b^{2} = r^{2} equiv x^{2} – 2ax + y^{2} – 2by = r^{2} – a^{2} – b^{2})
Comparing the RHS of the equation above with the equation rewritten as (x^{2} + y^{2} – 8x – 2y = -1), we have
(-2a = -8; -2b = -2 implies a = 4, b = 1)
(therefore r^{2} – 4^{2} – 1^{2} = -1 implies r^{2} = -1 + 16 + 1 = 16)
(r = sqrt{16} = 4)