A circle with centre (4,5) passes through the y-intercept of the line 5x - 2y + 6 = 0. Find its equation.
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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
\(x^{2} + y^{2} + 8x - 10y + 21 = 0\)
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
\(x^{2} + y^{2} + 8x - 10y + 21 = 0\)