A circle with centre (4,5) passes through the y-intercept of the line 5x – 2y + 6 = 0. Find its equation.
-
A.
(x^{2} + y^{2} + 8x – 10y + 21 = 0) -
B.
(x^{2} + y^{2} + 8x – 10y – 21 = 0) -
C.
(x^{2} + y^{2} – 8x – 10y – 21 = 0) -
D.
(x^{2} + y^{2} – 8x – 10y + 21 = 0)
Correct Answer: Option D
Explanation
On the y-intercept, x=0. At this point, y = 5(0) – 2y = -6; y = 3, so the y- intercept has coordinates (0,3).
The equation of the circle is given as ((x-a)^{2} + (y-b)^{2} = r^{2}), where (a,b) is the centre and r is the radius.
The radius of the circle through the y- intercept = (sqrt{(4-0)^{2} + (5-3)^{2}} = sqrt{20})
The equation of the circle is ((x-4)^{2} + (y-5)^{2} = 20); expanding, we have
(x^{2} + y^{2} + 8x – 10y + 21 = 0)