Circles
Circle Inscribed in Sector
Grade 11
Question:
<p>Let a circle be inscribed in the quadrant of a circle of diameter 4.</p><p><strong>Statement-1:</strong> The radius of inscribed circle is the positive root of the equation \(r^2 + 4r - 4 = 0\)</p><p><strong>Statement-2:</strong> Distance between their centres = \(2\) (radius of circle inscribed).</p>
<p>(A) Statement-1 is true, Statement-2 is true and Statement-2 is correct explanation for Statement-1</p>
<p>(B) Statement-1 is true, Statement-2 is true and Statement-2 is not correct explanation for Statement-1</p>
<p>(C) Statement-1 is true, Statement-2 is false</p>
<p>(D) Statement-1 is false, Statement-2 is true</p>
Step-by-Step Solution
Key Concept: The inscribed circle touches two perpendicular radii and the arc; the distance relationship arises from tangency conditions.
<p>A circle inscribed in a quadrant of radius 2 (diameter 4) touches both the radii and the arc. If the inscribed circle has radius \(r\) and centre at distance \(d\) from the vertex of the quadrant, then \(d = r\sqrt{2}\) (by the 45° geometry). The distance from vertex to arc centre is 2. By the tangency condition: \(d + r = 2\), giving \(r\sqrt{2} + r = 2\), so \(r(\sqrt{2}+1) = 2\). This leads to \(r^2 + 4r - 4 = 0\) when manipulated. Statement-2 correctly relates the geometry.</p>
Correct Answer: A