The line 2x - y + 1 = 0 is a tangent to the c | vedclass

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Question

(A) Let the centre of the circle be $O(h, k)$.Since the centre lies on the line $x - 2y = 4$,we have $h - 2k = 4$,which implies $k = \frac{h - 4}{2}$.

Vedclass · Accepted Answer

$3 \sqrt{5}$

Vedclass · Answer

(A) Let the centre of the circle be $O(h, k)$.Since the centre lies on the line $x - 2y = 4$,we have $h - 2k = 4$,which implies $k = \frac{h - 4}{2}$.So,the centre is $O(h, \frac{h - 4}{2})$.The line $2x - y + 1 = 0$ is a tangent at $A(2, 5)$. The radius $OA$ is perpendicular to the tangent.The slope of the tangent is $m_1 = 2$.The slope of the radius $OA$ is $m_2 = \frac{\frac{h - 4}{2} - 5}{h - 2} = \frac{h - 4 - 10}{2(h - 2)} = \frac{h - 14}{2(h - 2)}$.Since $m_1 	imes m_2 = -1$,we have $2 	imes \frac{h - 14}{2(h - 2)} = -1$.$\frac{h - 14}{h - 2} = -1 \implies h - 14 = -h + 2 \implies 2h = 16 \implies h = 8$.Then $k = \frac{8 - 4}{2} = 2$.The centre is $(8, 2)$.The radius $r$ is the distance between $(8, 2)$ and $(2, 5)$:$r = \sqrt{(8 - 2)^2 + (2 - 5)^2} = \sqrt{6^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45} = 3 \sqrt{5}$.

The line $2x - y + 1 = 0$ is a tangent to the circle at the point $(2, 5)$ and the centre of the circle lies on the line $x - 2y = 4$. Then,the radius of the circle is

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