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

Question

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. \(x^{2}+y^{2}-8 x-10 y+21=0\)

Answer 1

Correct Answer:

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
\[
x^{2}+y^{2}+8 x-10 y+21=0
\]