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A circle with centre $(4,5)$ passes through the y-intercept of the line $5 x-2 y+6=0$. Find its equation.

A. $x^{2}+y^{2}+8 x-10 y+21=0$
B. $x^{2}+y^{2}+8 x-10 y-21=0$
C. $x^{2}+y^{2}-8 x-10 y-21=0$
D. $x^{2}+y^{2}-8 x-10 y+21=0$
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Option D

Explanation
On the y-intercept, $x=0$. At this point, $y=5(0)-2 y=-6 ; y=3$, 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
$x^{2}+y^{2}+8 x-10 y+21=0$

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