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10-1.Circle and System of Circles
normal
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
A
$3 \sqrt{5}$
B
$5 \sqrt{3}$
C
$2 \sqrt{5}$
D
$5 \sqrt{2}$
Solution
$2x – y + 1 = 0$ is tangent
slope of line $OA = – \frac{1}{2}$
equation $(y – 5) = – \frac{1}{2}\,(x – 2)$
$2y – 10 = – x + 2$
$x + 2y = 12$
$\therefore$ 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 $OA =\sqrt {{{(8 – 2)}^2} + {{(2 – 5)}^2}} = \sqrt {36 + 9} = \sqrt{45} =3 \sqrt{5}$
Standard 11
Mathematics
