First slide

Definition of a circle

Question

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

NA

A

$3 \sqrt{5}$
B

$5 \sqrt{3}$
C

$2 \sqrt{5}$
D

$5 \sqrt{2}$

Solution

2x - y + I = 0 is a tangent.
$Slope of line OA = - \frac{1}{2}$

The equation of OA is
$(y - 5) = - \frac{1}{2} (x - 2) or x + 2 y = 12 Therefore, intersection with x - 2 y = 4 will give the coordinates of center as (8, 2). Hence, r = OA = \sqrt{(8 - 2)^{2} + (2 - 5)^{2}} = 3 \sqrt{5}$

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