<p>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:</p>
Step-by-Step Solution
Key Concept: Since a line is tangent to a circle at a point, the radius to that point is perpendicular to the tangent line. Use this perpendicularity condition and the constraint that the centre lies on a given line to find the centre, then calculate the radius as the distance from centre to the tangent point.
<p><strong>Step 1: Find the slope of the tangent line.</strong></p><p>The tangent line is $2x - y + 1 = 0$, which can be written as $y = 2x + 1$. The slope of the tangent is $m_{\text{tangent}} = 2$.</p><p><strong>Step 2: Find the slope of the radius.</strong></p><p>Since the radius at the point of tangency is perpendicular to the tangent line, the slope of the radius is $m_{\text{radius}} = -\frac{1}{2}$ (negative reciprocal).</p><p><strong>Step 3: Write the equation of the line through the centre.</strong></p><p>The radius passes through the point $(2, 5)$ with slope $-\frac{1}{2}$. The equation of this line is:</p><p>$$y - 5 = -\frac{1}{2}(x - 2)$$</p><p>$$y - 5 = -\frac{1}{2}x + 1$$</p><p>$$y = -\frac{1}{2}x + 6$$</p><p>Or: $2y = -x + 12$, i.e., $x + 2y = 12$</p><p><strong>Step 4: Find the centre of the circle.</strong></p><p>The centre lies on both $x + 2y = 12$ (from Step 3) and $x - 2y = 4$ (given). Solving simultaneously:</p><p>Adding the equations: $2x = 16 \Rightarrow x = 8$</p><p>Substituting into $x - 2y = 4$: $8 - 2y = 4 \Rightarrow 2y = 4 \Rightarrow y = 2$</p><p>So the centre is $(8, 2)$.</p><p><strong>Step 5: Calculate the radius.</strong></p><p>The radius is the distance from the centre $(8, 2)$ to the point of tangency $(2, 5)$:</p><p>$$r = \sqrt{(8-2)^2 + (2-5)^2} = \sqrt{36 + 9} = \sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5}$$</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A