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

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Question

$2x - y + 1 = 0$ is tangent

slope of line $OA = - \frac{1}{2}$

equation $(y - 5) = - \frac{1}{2}\,(x - 2)$

$2y - 10 = - x + 2$

$x + 2y = 1

Vedclass · Accepted Answer

$3 \sqrt{5}$

Vedclass · Answer

$2x - y + 1 = 0$ is tangent

slope of line $OA = - \frac{1}{2}$

equation $(y - 5) = - \frac{1}{2}\,(x - 2)$

$2y - 10 = - x + 2$

$x + 2y = 12$

$	herefore$ intersection with $x - 2y = y$ will give coordinates of centre

$x + 2y = 12$

$- x - 2y = 4$

                        +

________________

$4y = 8\Rightarrow y = 2$

$x - 4 = 4\Rightarrow x = 8$

$C = (8, 2)$

distance $OA =\sqrt {{{(8 - 2)}^2} + {{(2 - 5)}^2}}  = \sqrt {36 + 9} = \sqrt{45} =3 \sqrt{5}$

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

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